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Collected inequality from the other sites
can_hang2007
 This topic, I will collect the inequalities from Mathlinks forum with their solutions.
Problem 1. (Shaam) Given that a,b,c \geq 0, and a + b + c = 3, prove that
a+ab+2abc \le \frac{9}{2}.
Link: http://www.mathli...39#1440739
Solution 1. (can_hang2007) Replacing b=3-a-c, then we have to prove
a+a(3-a-c)+2ac(3-a-c) \le \frac{9}{2},
or equivalently,
f(a)=(2c+1)a^2+(2c^2-5c-4)a+\frac{9}{2} \ge 0.
We see that f(a) is a quadratic polynomial of a with the highest coefficient is positive. Moreover, its disciminant is
\Delta= (2c^2-5c-4)^2-18(2c+1) =(2c-1)^2(c^2-4c-2) \le 0, as 0 \le c \le 3.
Therefore, f(a) \ge 0 and our proof is completed. Equality holds if and only if a=\frac{3}{2},b=1,c=\frac{1}{2}. Smile

Solution 2 (Shaam). Applying AM - GM Inequality, we see that
a+ab+2abc=a+2ab\left( c+\frac{1}{2}\right)  \le a+2a\left( \frac{b+c+\frac{1}{2}}{2}\right)^2 =a+2 a\left( \frac{7-2a}{4}\right)^2.
It suffices to prove that
a+2 a\left( \frac{7-2a}{4}\right)^2 \le \frac{9}{2},
which is equivalent to (4-a)(3-2a)^2 \ge 0, which is true as 4>a \ge 0.

Problem 2. (ductrung, a present to manlio) For all nonnegative real numbers a, b and c, no two of which are zero,
\frac {a(b + c)(a^2 - bc)}{a^2 + bc} + \frac {b(c + a)(b^2 - ca)}{b^2 + ca} + \frac {c(a + b)(c^2 - ab)}{c^2 + ab}\ge 0.
Link: http://www.mathli...p?t=264746
Solution. (can_hang2007) Setting A = LHS, then we see that
A = \sum \frac {a^3(b + c)}{a^2 + bc} - abc\sum \frac {b + c}{a^2 + bc} \ge \sum \frac {a^3(b + c)}{a^2 + bc} - abc\sum \frac {1}{a}
= \sum \frac {a^3(b + c)}{a^2 + bc} - \sum \frac {a(b + c)}{2} = \frac {1}{2} \sum \frac {a(b + c)(a^2 - bc)}{a^2 + bc} = \frac {1}{2}A,
which shows that A \ge 0. Smile

Problem 3. (Vasile Cirtoaje) Let a,b,c be nonnegative real numbers, no two of which are zero. Prove that
a) \frac 1{2a^2 + bc} + \frac 1{2b^2 + ca} + \frac 1{2c^2 + ab}\ge \frac 1{ab + bc + ca} + \frac 2{a^2 + b^2 + c^2};
b) \frac 2{3}(\frac 1{a^2 + bc} + \frac 1{b^2 + ca} + \frac 1{c^2 + ab})\ge \frac 1{ab + bc + ca} + \frac 2{a^2 + b^2 + c^2};
c) 2(\frac 1{a^2 + 8 bc} + \frac 1{b^2 + 8ca} + \frac 1{c^2 + 8ab})\ge \frac 1{ab + bc + ca} + \frac 1{a^2 + b^2 + c^2};
d) (Darij Grinberg) \frac 1{a^2 + 2 bc} + \frac 1{b^2 + 2ca} + \frac 1{c^2 + 2ab}\ge \frac 2{ab + bc + ca} + \frac 1{a^2 + b^2 + c^2};
e) \frac 5{3}(\frac 1{4a^2 + bc} + \frac 1{4b^2 + ca} + \frac 1{4c^2 + ab})\ge \frac 2{ab + bc + ca} + \frac 1{a^2 + b^2 + c^2}.
Link: http://www.mathli...p?t=263911
Solution. (can_hang2007)
a) Assume that c = \min\{a,b,c\}, then the Cauchy Schwarz Inequality yields
\frac {1}{2a^2 + bc} + \frac {1}{2b^2 + ca} \ge \frac {4}{2(a^2 + b^2) + c(a + b)},
then we just need to prove that
\frac {4}{2(a^2 + b^2) + c(a + b)} + \frac {1}{ab + 2c^2} \ge \frac {1}{ab + bc + ca} + \frac {2}{a^2 + b^2 + c^2},
or equivalently
\frac {c(a + b - 2c)}{(ab + 2c^2)(ab + bc + ca)} \ge \frac {2c(a + b - 2c)}{(a^2 + b^2 + c^2)(2a^2 + 2b^2 + ac + bc)} ,
that is
(a^2 + b^2 + c^2)(2a^2 + 2b^2 + ac + bc) \ge 2(ab + 2c^2)(ab + bc + ca),
which is true since
a^2 + b^2 + c^2 \ge ab + bc + ca and 2a^2 + 2b^2 + ac + bc \ge 2(ab + 2c^2).
This completes our proof. Smile
b) Rewrite our inequality as:
\sum \frac {1}{a^2 + bc} \ge \frac {3(a + b + c)^2}{2(a^2 + b^2 + c^2)(ab + bc + ca)}.
We will consider 2 cases:
Case 1. a^2 + b^2 + c^2 \le 2(ab + bc + ca), then applying Cauchy Schwarz Inequality, we can reduce our inequality to
\frac {6}{a^2 + b^2 + c^2 + ab + bc + ca} \ge \frac {(a + b + c)^2}{(a^2 + b^2 + c^2)(ab + bc + ca)},
(a^2 + b^2 + c^2 - ab - bc - ca)(2ab + 2bc + 2ca - a^2 - b^2 - c^2) \ge 0, which is true.
Case 2. a^2 + b^2 + c^2 \ge 2(ab + bc + ca), then (a + b + c)^2 \le 2(a^2 + b^2 + c^2), which yields that
\frac {3(a + b + c)^2}{2(a^2 + b^2 + c^2)(ab + bc + ca)} \le \frac {3}{ab + bc + ca},
and we just need to prove that
\frac {1}{a^2 + bc} + \frac {1}{b^2 + ca} + \frac {1}{c^2 + ab} \ge \frac {3}{ab + bc + ca},
which is just a very known inequality. Smile
c) Replacing a,b,c by \frac {1}{a},\frac {1}{b},\frac {1}{c} respectively, we can rewrite our inequality as
4(a + b + c)\left( \frac {a}{8a^2 + bc} + \frac {b}{8b^2 + ca} + \frac {c}{8c^2 + ab}\right) \ge 2 + \frac {2abc(a + b + c)}{a^2b^2 + b^2c^2 + c^2a^2}.
Now, assume that c = \min\{a,b,c\}, then we have the following estimations:
\frac {a(4a + 4b + c)}{8a^2 + bc} + \frac {b(4a + 4b + c)}{8b^2 + ca} - 2 = \frac {(a - b)^2(32ab - 12ac - 12bc + c^2)}{(8a^2 + bc)(8b^2 + ca)} \ge 0,
and
\frac {2abc(a + b + c)}{a^2b^2 + b^2c^2 + c^2a^2} \le \frac {2c(a + b + c)}{ab + 2c^2}.
With these estimations, we can reduce our inequality to
\frac {3ac}{8a^2 + bc} + \frac {3bc}{8b^2 + ca} + \frac {4c(a + b + c)}{8c^2 + ab} \ge \frac {2c(a + b + c)}{ab + 2c^2},
or
\frac {3a}{8a^2 + bc} + \frac {3b}{8b^2 + ca} \ge \frac {2(a + b + c)(4c^2 - ab)}{(ab + 2c^2)(ab + 8c^2)}.
According to Cauchy Schwarz Inequality, we have
\frac {3a}{8a^2 + bc} + \frac {3b}{8b^2 + ca} \ge \frac {12}{8(a + b) + c\left( \frac {a}{b} + \frac {b}{a}\right)}.
It suffices to show that
\frac {6}{8(a + b) + c\left( \frac {a}{b} + \frac {b}{a}\right)}\ge \frac {(a + b + c)(4c^2 - ab)}{(ab + 2c^2)(ab + 8c^2)}.
If 4c^2 \le ab, then it is trivial. Otherwise, we have
a + b \le c + \frac {ab}{c}, and \frac {a}{b} + \frac {b}{a} \le \frac {ab}{c^2} + \frac {c^2}{ab}.
We need to prove
\frac {6}{8\left( c + \frac {ab}{c}\right) + c\left( \frac {ab}{c^2} + \frac {c^2}{ab}\right)} \ge \frac {\left( 2c + \frac {ab}{c}\right) (4c^2 - ab)}{(ab + 2c^2)(ab + 8c^2)},
which is, after expanding, equivalent to
\frac {(9ab - 4c^2)(ab - c^2)^2}{c(ab + 8c^2)(c^4 + 8abc^2 + 9a^2b^2)} \ge 0, which is true as c = \min\{a,b,c\}.
Our proof is completed. Smile

d) It is easy to see that the inequality is equivalent to (a - b)^2(b - c)^2(c - a)^2 \ge 0 (trivial). Smile

e) Assume that c = \min\{a,b,c\}, then we have the following estimations:
\frac {1}{4a^2 + bc} + \frac {1}{4b^2 + ca} - \frac {4}{8ab + ac + bc} = \frac {(a - b)^2(32ab - 12ac - 12bc + c^2)}{(4a^2 + bc)(4b^2 + ca)(8ab + ac + bc)} \ge 0,
and
\frac {1}{a^2 + b^2 + c^2} \le \frac {1}{2ab + c^2}.
Using these, we can reduce our inequality to
\frac {20}{8ab + ac + bc} + \frac {5}{ab + 4c^2} \ge \frac {6}{ab + ac + bc} + \frac {3}{2ab + c^2}.
Denote x = a + b \ge 2\sqrt {ab}, then this inequality can be rewritten as
f(x) = \frac {20}{cx + 8ab} - \frac {6}{cx + ab} + \frac {5}{ab + 4c^2} - \frac {3}{2ab + c^2} \ge 0.
We have
f'(x) = \frac {6c}{(cx + ab)^2} - \frac {20}{(cx + 8ab)^2} \ge \frac { 20c}{(cx + ab)(cx + 8ab)} - \frac {20c}{(cx + 8ab)^2} = \frac {140abc}{(cx + ab)(cx + 8ab)^2} \ge 0.
This shows that f(x) is increasing, and we just need to prove that f(2\sqrt {ab}) \ge 0, which is equivalent to
\frac {7c(13t^2 + 6tc + 8c^2)(t - c)^2}{t(t + 2c)(4t + c)(2t^2 + c^2)(t^2 + 4c^2)} \ge 0, (where t = \sqrt {ab}).
This is obviously nonnegative, so our proof is completed. Smile
Sửa bởi can_hang2007 vào lúc 26-03-2009 14:08
The love makes us stronger!
 
can_hang2007
 Today, I will continue with the two new inequalities of Vasile Cirtoaje as follow:

Problem 4. (Vasile Cirtoaje) Find the smallest value of and such that the inequalities below holds for any nonnegative real numbers , no two of which are zero:
a) \frac{a(b+c)}{a^2+2bc} +\frac{b(c+a)}{b^2+2ca}+\frac{c(a+b)}{c^2+2ab} -2+k_1 \left( \frac{a^2+b^2+c^2}{ab+bc+ca} -1\right) \ge 0;
b) \frac{a(b+c)}{2a^2+bc} +\frac{b(c+a)}{2b^2+ca}+\frac{c(a+b)}{2c^2+ab} -2+k_2 \left( \frac{a^2+b^2+c^2}{ab+bc+ca} -1\right) \ge 0.
Link: http://www.mathli...p?t=265697
Hint. (can_hang2007) The answer is k_1=\frac{1}{\sqrt{2}}, and k_2=\frac{32}{27}.
I will just write the hint, the detail proofs will be left to the readers Smile
a) Since \frac{a(b+c)}{a^2+2bc} -\frac{a(b+c)}{ab+bc+ca} =-\frac{a(b+c)(a-b)(a-c)}{(ab+bc+ca)(a^2+2bc)}, and
\frac{a^2+b^2+c^2}{ab+bc+ca} -1 =\frac{(a-b)(a-c)+(b-c)(b-a)+(c-a)(c-b)}{ab+bc+ca},
we can rewrite our inequality as
\sum (a-b)(a-c)\left( \frac{1}{\sqrt{2}} -\frac{a(b+c)}{a^2+2bc}\right) \ge 0.
To use the Vornicu Schur method, the terms x,y,z of \sum x(a-b)(a-c) must be nonnegative, but unfortunately, the inequality \frac{1}{\sqrt{2}} -\frac{a(b+c)}{a^2+2bc} \ge 0, does not always hold. Therefore, we can't use directly the Vornicu Schur technique. However, with note at the trivial identity
(b-c)^2(a-b)(a-c)+(c-a)^2(b-c)(b-a)+(a-b)^2(c-a)(c-b) =0,
we can rewrite our inequality in the form
\sum (a-b)(a-c) \left( \frac{1}{\sqrt{2}} -\frac{a(b+c)}{a^2+2bc}+\frac{(b-c)^2}{ab+bc+ca} \right) \ge 0,
It is easy to prove that
 \frac{1}{\sqrt{2}} -\frac{a(b+c)}{a^2+2bc}+\frac{(b-c)^2}{ab+bc+ca} \right) \ge \frac{1}{\sqrt{2}} -\frac{a(b+c)}{a^2+\frac{(b+c)^2}{2}} \ge 0.
Therefore, we can use the Vornicu Schur technique for x(a-b)(a-c)+y(b-c)(b-a)+z(c-a)(c-b) \ge 0, where
x= \frac{1}{\sqrt{2}} -\frac{a(b+c)}{a^2+2bc}+\frac{(b-c)^2}{ab+bc+ca} \ge 0.
Now, assume that a \ge b \ge c, we can prove that z \ge y (the readers can try this) and with the theorem of Vornicu Schur, we can get the result.
Equality holds for a=b=c and again for a=\sqrt{2}b=\sqrt{2}c (with its cyclic permutations). Smile

b) By the same idea with a), we can rewrite our inequality as
\sum (a-b)(a-c) \left( \frac{32}{27}\cdot \frac{a+b+c}{ab+bc+ca}+\frac{(a+b+c)(b-c)^2}{2(ab+bc+ca)} -\frac{b+c}{2a^2+bc}\right) \ge 0.
This inequality has the form x(a-b)(a-c)+y(b-c)(b-a)+z(c-a)(c-b) \ge 0, with
x = \frac{32}{27}\cdot \frac{a+b+c}{ab+bc+ca}+\frac{(a+b+c)(b-c)^2}{2(ab+bc+ca)} -\frac{b+c}{2a^2+bc} \ge 0.
Now, assume that a \ge b \ge c, we can prove that ax \ge by (the readers can try this, too) and with the theorem of Vornicu Schur, we can get the result.
Equality holds for a=b=c and again for a=b=4c (with its cyclic permutations). Smile
Sửa bởi can_hang2007 vào lúc 23-03-2009 04:21
The love makes us stronger!
 
daogiauvang
 Lời giải dùng Schur-Vonicur rất thú vị. Nhưng hằng đẳng thức bổ sung rất đặc biệt.

----

Nhắc nhở: daogiauvang, chú ý là bạn đang ở các Box English đấy nhé!
Sửa bởi Vnkvant vào lúc 23-03-2009 08:57
If I feel unhappy , I do mathematics to become happy , If I feel happy , I do mathematics to keep happy

http://360.yahoo....ntethegioi
 
http://tpu.ru
can_hang2007
 Continue:
Problem 5. (Ductrung) Let a,b,c be nonnegative real numbers, no two of which are zero. Prove that
\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2}+\frac{1}{(a+b)^2} \ge \frac{(a+b+c)^4}{4(a^2+b^2+c^2)(ab+bc+ca)^2}.
Link: http://www.mathli...p?t=266180
Proof. (can_hang2007) This inequality follows from
(a^2 + b^2 + c^2) \left( \frac {1}{(b + c)^2} + \frac {1}{(c + a)^2} + \frac {1}{(a + b)^2}\right) \ge \left( \frac {a}{b + c} + \frac {b}{c + a} + \frac {c}{a + b}\right)^2,
and
\frac {a}{b + c} + \frac {b}{c + a} + \frac {c}{a + b} \ge \frac {(a + b + c)^2}{2(ab + bc + ca)}.
Smile
The love makes us stronger!
 
can_hang2007
 Continue...
Problem 6. (secrets) Let a,b,c be nonnegative real numbers, no two of which are zero. Prove that
\sqrt{\frac{b^2-bc+c^2}{a^2+bc}}+\sqrt{\frac{c^2-ca+a^2}{b^2+ca}}+\sqrt{\frac{a^2-ab+b^2}{c^2+ab}} +\frac{2(ab+bc+ca)}{a^2+b^2+c^2} \ge 4.
Link: http://www.mathli...p?t=266228
Proof. (can_hang2007) This inequality follows from applying AM-GM as follow:
\sqrt {\frac {b^2 - bc + c^2}{a^2 + bc}} \ge \frac {2(b^2 - bc + c^2)}{(a^2 + bc) + (b^2 - bc + c^2)} = \frac {2(b^2 - bc + c^2)}{a^2 + b^2 + c^2},
;) Smile
The love makes us stronger!
 
can_hang2007
 Continue...
Problem 7. (Ductrung) For all nonnegative real numbers a,b and c, no two of which are zero,
\frac {1}{(a + b)^2} + \frac {1}{(b + c)^2} + \frac {1}{(c + a)^2} \ge \frac {3\sqrt {3abc(a + b + c)}(a + b + c)^2}{4(ab + bc + ca)^3}.
Link: http://www.mathli...p?t=266355
Proof. (can_hang2007) Replacing a,b,c by \frac{1}{a},\frac{1}{b},\frac{1}{c} respectively, we have to prove that
\sum\frac{a^2b^2}{(a+b)^2} \ge \frac{3\sqrt{3(ab+bc+ca)}(ab+bc+ca)^2}{4(a+b+c)^3}.
Now, using Cauchy Schwarz Inequality, we have
\sum \frac{a^2b^2}{(a+b)^2} \ge \frac{(ab+bc+ca)^2}{(a+b)^2+(b+c)^2+(c+a)^2} =\frac{(ab+bc+ca)^2}{2(a^2+b^2+c^2+ab+bc+ca)}.
It suffices to prove that
\frac{(ab+bc+ca)^2}{2(a^2+b^2+c^2+ab+bc+ca)} \ge \frac{3\sqrt{3(ab+bc+ca)}(ab+bc+ca)^2}{4(a+b+c)^3}
or equivalently,
2(a+b+c)^3 \ge 3\sqrt{3(ab+bc+ca)}(a^2+b^2+c^2+ab+bc+ca),
that is
4(a+b+c)^6 \ge 27(ab+bc+ca)(a^2+b^2+c^2+ab+bc+ca)^2
By AM-GM, we see that
27(ab+bc+ca)(a^2+b^2+c^2+ab+bc+ca)^2 \le \frac{1}{2} \left( 2(ab+bc+ca)+(a^2+b^2+c^2+ab+bc+ca)+(a^2+b^2+c^2+ab+bc+ca)\right)^3 =4(a+b+c)^6.
Therefore, our proof is completed. Smile
The love makes us stronger!
 
can_hang2007
 Continue...
Problem 8. (Ductrung, a stronger version of problem 5) Let a,b,c be nonnegative real numbers, no two of which are zero. Prove that
\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2}+\frac{1}{(a+b)^2} \ge \frac{3(a+b+c)^2}{8(ab+bc+ca)}\left( \frac{1}{ab+bc+ca}+\frac{1}{a^2+b^2+c^2}\right).
Link: http://www.mathli...p?t=266180
Proof. (can_hang2007) See at my attachment. Smile
Sửa bởi can_hang2007 vào lúc 26-03-2009 12:56
The love makes us stronger!
 
can_hang2007
 Problem 9. (Vasile Cirtoaje, after a Phan Thanh Nam's problem) Let x,y,z be positive real numbers with sum 1. Prove that
\sqrt{x+\frac{(y-z)^2}{p}} + \sqrt{y+\frac{(z-x)^2}{p}} + \sqrt{z+\frac{(x-y)^2}{p}}  \le \sqrt{3}, with p=4+\sqrt{21}.
Link: http://www.mathli...hp?t=18182
Proof. (Vasile Cirtoaje, using Darkseer way) Substitue a=\frac{(y-z)^2}{p}, and b,c are similar. Then, squaring both sides, we can rewrite our inequality as
\sum[(\sqrt {y + b} - \sqrt {z + c})^2 - 3a]\geq0.
It suffices to show that
(\sqrt {y + b} - \sqrt {z + c})^2\geq 3a,
which reduces to
(\frac {3x + p - 1}{(\sqrt {y + b} + \sqrt {z + c}})}^2\geq 3p.
Since (\sqrt {y + b} + \sqrt {z + c})^2\leq2(y + z + b + c), it suffices to show that
(3x + p - 1)^2\geq 6p(1 - x + b + c),
that is
(3x + p - 1)^2 - 6p(1 - x) - 6(5x^2 - 4x + 1) + 12yz\geq 0.
It suffices to prove that
f(x) = (3x + p - 1)^2 - 6p(1 - x) - 6(5x^2 - 4x + 1) \geq 0.
Since f(x) is concave, it suffices to show that f(0)\geq 0 and f(1)\geq 0.
These inequalities hold for p\geq 4 + \sqrt {21}. Smile
Sửa bởi can_hang2007 vào lúc 26-03-2009 13:18
The love makes us stronger!
 
can_hang2007
 Problem 10. (Vasile Cirtoaje) Let a,b,c be positive real numbers. Prove that
\frac{a^3}{2a^2+b^2}+\frac{b^3}{2b^2+c^2}+\frac{c^3}{2c^2+a^2} \ge \frac{a+b+c}{3}.
Link: http://www.mathli...77#1448077
Proof 1. (Michael Rozenberg) We have
\sum\frac {a^3}{2a^2 + b^2} \ge \frac {a + b + c}{3}\Leftrightarrow\sum(\frac {a^3}{2a^2 + b^2} - \frac {a}{3})\geq0\Leftrightarrow
\Leftrightarrow\sum(\frac {(a - b)(a^2 + ab)}{2a^2 + b^2} - \frac {2}{3}(a - b))\Leftrightarrow\sum\frac {(2b - a)(a - b)^2}{2a^2 + b^2}\geq0.
Hence, if \{\frac {a}{b},\frac {b}{c},\frac {c}{a}\}\subset(0,2] our inequality is proved.
1) a\geq b\geq c.
Since \frac {c}{a}\in(0,2], remain to prove our inequality for a\geq2b or b\geq2c.
If a\geq b\geq2c then \frac {a^3}{2a^2 + b^2}\geq\frac {a}{2} - \frac {b}{6} and \frac {b^3}{2b^2 + c^2}\geq\frac {b}{2} - \frac {c}{9}.
Hence, \sum\frac {a^3}{2a^2 + b^2}\geq\frac {a}{2} - \frac {b}{6} + \frac {b}{2} - \frac {c}{9} + \frac {c^3}{2c^2 + a^2} = \frac {a}{2} + \frac {b}{3} - \frac {c}{9} + \frac {c^3}{2c^2 + a^2}.
But \frac {a}{2} + \frac {b}{3} - \frac {c}{9} + \frac {c^3}{2c^2 + a^2}\geq\frac {a + b + c}{3}\Leftrightarrow\frac {c^3}{2c^2 + a^2}\geq\frac {4c}{9} - \frac {a}{6}\Leftrightarrow
\Leftrightarrow2t^3 + 6t^2 - 8t + 3\geq0, where t = \frac {c}{a}. This is true, since 6t^2 - 8t + 3 > 0.
If a\geq2b\geq2c then \frac {a^3}{2a^2 + b^2}\geq\frac {a}{2} - \frac {b}{9} and \frac {b^3}{2b^2 + c^2}\geq\frac {4b}{9} - \frac {c}{9}.
Hence, \sum\frac {a^3}{2a^2 + b^2}\geq\frac {a}{2} - \frac {b}{9} + \frac {4b}{9} - \frac {c}{9} + \frac {c^3}{2c^2 + a^2} =
= \frac {a}{2} + \frac {b}{3} - \frac {c}{9} + \frac {c^3}{2c^2 + a^2}\geq\frac {a + b + c}{3}. The last is previously proved.
2) a\geq c\geq b. Here \{\frac {b}{c},\frac {c}{a}\}\subset(0,2]. Hence, remain only the case a\geq2b.
\frac {a^3}{2a^2 + b^2}\geq\frac {22a - 4b}{45}\Leftrightarrow(a - 2b)(a^2 + 10ab - 2b^2)\geq0 and
\frac {b^3}{2b^2 + c^2}\geq\frac {19b - 5.8c}{45}\Leftrightarrow 7t^3 + 11.6t^2 - 19t + 5.8\geq0, where t = \frac {b}{c}.
If f(t) = 7t^3 + 11.6t^2 - 19t + 5.8 then
f'(t) = 21t^2 + 23.2t - 19, t_{min} = \frac { - 11.6 + \sqrt {11.6^2 + 21\cdot19}}{21} = 0.5475...
Hence, f(t)\geq f(t_{min}) = 0.023488... > 0. Pfft
Hence, \sum\frac {a^3}{2a^2 + b^2}\geq\frac {22a - 4b}{45} + \frac {19b - 5.8c}{45} + \frac {c^3}{2c^2 + a^2} =
= \frac {22a}{45} + \frac {b}{3} - \frac {5.8c}{45} + \frac {c^3}{2c^2 + a^2}.
But \frac {22a}{45} + \frac {b}{3} - \frac {5.8c}{45} + \frac {c^3}{2c^2 + a^2}\geq\frac {a + b + c}{3}\Leftrightarrow\frac {c^3}{2c^2 + a^2}\geq\frac {20.8c - 7a}{45}\Leftrightarrow
\Leftrightarrow3.4t^3 + 14t^2 - 20.8t + 7\geq0, where t = \frac {c}{a}.
If f(t) = 3.4t^3 + 14t^2 - 20.8t + 7 then f'(t) = 10.2t^2 + 28t - 20.8 and
t_{min} = \frac { - 14 + \sqrt {14^2 + 10.2\cdot20.8}}{10.2} = 0.608...
Id est, f(t)\geq f(t_{min}) = 0.293... > 0,
and our inequality is proved. Smile

Proof 2. (can_hang2007) See at my attachment. Smile
Sửa bởi can_hang2007 vào lúc 26-03-2009 13:31
The love makes us stronger!
 
can_hang2007
 Problem 11. (can_hang2007) Let a,b,c be nonnegative real numbers. Prove that
\underset{cyc}{\sum}\frac {1}{\sqrt {a^{2} + bc}}\le \underset{cyc}{\sum}\frac {\sqrt {2}}{a + b}.
Link: http://www.mathli...p;start=20
Proof. (can_hang2007)
By the Cauchy schwarz Inequality, we have
\left(\sum \frac {1}{\sqrt {a^2 + bc}}\right)^2 \le \left(\sum \frac {1}{(a + b)(a + c)}\right)\left(\sum \frac {(a + b)(a + c)}{a^2 + bc}\right)} = \frac {2\sum a}{(a + b)(b + c)(c + a)}\left(\sum \frac {a(b + c)}{a^2 + bc} + 3\right).
It suffices to show that
\frac {2\sum a}{(a + b)(b + c)(c + a)}\left(\sum \frac {a(b + c)}{a^2 + bc} + 3\right) \le 2\left( \sum \frac {1}{a + b}\right)^2,
\sum \frac {a(b + c)}{a^2 + bc} + 3 \le \frac {\left( \sum a^2 + 3\sum ab)^2}{(a + b)(b + c)(c + a)\sum a},
\sum \frac {a(b + c)}{a^2 + bc} - 3 \le \frac { \sum a^4 - \sum a^2b^2}{(a + b)(b + c)(c + a)\sum a},
\sum (a - b)(a - c)\left(\frac {1}{a^2 + bc} + \frac {1}{(b + c)(a + b + c)}\right) \ge 0.
Due to symmetry, we may assume a \ge b \ge c, since a - c \ge \frac {a}{b}(b - c). It suffices to show that
a\left(\frac {1}{a^2 + bc} + \frac {1}{(b + c)(a + b + c)}\right) \ge b\left(\frac {1}{b^2 + ca} + \frac {1}{(a + c)(a + b + c)}\right),
c(a^2 - b^2)[(a - b)^2 + ab + bc + ca] \ge 0, which is trivial.
Equality holds if and only if a = b = c. Smile
Sửa bởi can_hang2007 vào lúc 26-03-2009 13:52
The love makes us stronger!
 
can_hang2007
 Problem 12. (Dduclam) Let a,b,c be nonnegative real numbers, no two of which are zero. Prove that
\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \ge \frac{a}{\sqrt{a^2+3bc}}+\frac{b}{\sqrt{b^2+3ca}}+\frac{c}{\sqrt{c^2+3ab}}.
Link: http://www.mathli...p?t=266716
Proof. (can_hang2007) After using Cauchy Schwarz, we can see that the inequality follows from
\sum \frac {a}{b + c} \ge \sum \frac {a(b + c)}{a^2 + 3bc},
that is
\sum \frac {a(a^2 + bc - b^2 - c^2)}{(b + c)(a^2 + 3bc)} \ge 0,
or
\sum \frac {a^3(b + c) - a(b^3 + c^3)}{(b + c)^2(a^2 + 3bc)} \ge 0.
Without loss of generality, we can assume that a \ge b \ge c, then
a^3(b + c) - a(b^3 + c^3) \ge 0 \ge c^3(a + b) - c(a^3 + b^3),
and
\frac {1}{(b + c)^2(a^2 + 3bc)} \ge \frac {1}{(c + a)^2(b^2 + 3ca)} \ge \frac {1}{(a + b)^2(c^2 + 3ab)}.
It follows that
\frac {a^3(b + c) - a(b^3 + c^3)}{(b + c)^2(a^2 + 3bc)} \ge \frac {a^3(b + c) - a(b^3 + c^3)}{(c + a)^2(b^2 + 3ca)},
and
\frac {c^3(a + b) - c(a^3 + b^3)}{(a + b)^2(c^2 + 3ab)} \ge \frac {c^3(a + b) - c(a^3 + b^3)}{(c + a)^2(b^2 + 3ca)}.
Therefore
\sum \frac {a^3(b + c) - a(b^3 + c^3)}{(b + c)^2(a^2 + 3bc)} \ge \frac {\sum (a^3(b + c) - a(b^3 + c^3))}{(c + a)^2(b^2 + 3ca)} = 0.
Our proof is completed. Smile
The love makes us stronger!
 
can_hang2007
 Problem 13. (can_hang2007) Let a,b,c be nonnegative real numbers, no two of which are zero. Prove that
\sqrt[3]{\frac{a^2+bc}{b^2+c^2}}+\sqrt[3]{\frac{b^2+ca}{c^2+a^2}}+\sqrt[3]{\frac{c^2+ab}{a^2+b^2}} \ge 2+\frac{1}{\sqrt[3]{2}}.
Link: http://www.mathli...p;start=20
Proof. (can_hang2007) Assume that a \ge b \ge c, then we will show that
\sqrt [3]{\frac {a^2 + bc}{b^2 + c^2}} + \sqrt [3]{\frac {b^2 + ca}{c^2 + a^2}} \ge \max\left\{ 2, \sqrt [3]{\frac {4(a^2 + b^2)}{c^2 + ab}}\right\}.
Indeed, we have
\sqrt [3]{\frac {a^2 + bc}{b^2 + c^2}} + \sqrt [3]{\frac {b^2 + ca}{c^2 + a^2}} \ge 2\sqrt [6]{\frac {(a^2 + bc)(b^2 + ca)}{(b^2 + c^2)(a^2 + c^2)}} \ge 2\sqrt [6]{\frac {(a^2 + c^2)(b^2 + c^2)}{(b^2 + c^2)(a^2 + c^2)}} = 2.
So, to prove our claim, we need to show that
\sqrt [3]{\frac {a^2 + bc}{b^2 + c^2}} + \sqrt [3]{\frac {b^2 + ca}{c^2 + a^2}} \ge \sqrt [3]{\frac {4(a^2 + b^2)}{c^2 + ab}},
 \frac {a^2 + bc}{b^2 + c^2} + \frac {b^2 + ca}{c^2 + a^2} + 3\sqrt [3]{\frac {(a^2 + bc)(b^2 + ca)}{(a^2 + c^2)(b^2 + c^2)}} \left( \sqrt [3]{\frac {a^2 + bc}{b^2 + c^2}} + \sqrt [3]{\frac {b^2 + ca}{c^2 + a^2}}\right) \ge \frac {4(a^2 + b^2)}{c^2 + ab}.
We have
\frac {b^2 + ca}{b^2 + c^2} - \frac {a^2 + bc}{a^2 + c^2} = \frac {c(a - b)(a^2 + b^2 + c^2 + ab - ac - bc)}{(a^2 + c^2)(b^2 + c^2)} \ge 0.
Thus, by AM-GM Inequality, we have
3\sqrt [3]{\frac {(a^2 + bc)(b^2 + ca)}{(a^2 + c^2)(b^2 + c^2)}} \left( \sqrt [3]{\frac {a^2 + bc}{b^2 + c^2}} + \sqrt [3]{\frac {b^2 + ca}{c^2 + a^2}}\right) \ge 6\sqrt {\frac {(a^2 + bc)(b^2 + ca)}{(a^2 + c^2)(b^2 + c^2)}} \ge \frac {6(a^2 + bc)}{a^2 + c^2}.
It suffices for us to show that
\frac {a^2 + bc}{b^2 + c^2} + \frac {b^2 + ca}{c^2 + a^2} + \frac {6(a^2 + bc)}{a^2 + c^2} \ge \frac {4(a^2 + b^2)}{c^2 + ab},
which is equivalent to f(c) + g(c) \ge 0, where
f(c) = (a + 7b)c^5 + 3(a^2 - b^2)c^4 + 2(a + b)(a + 3b)bc^3 \ge 0,
g(c) = (a - b)(3b^3 + 2ab^2 + 4a^2b - 3a^3)c^2 + (b^2a^3 + 6b^4a + a^2b^3)c + ab(a - b)^4.
If 3b^3 + 2ab^2 + 4a^2b - 3a^3 \ge 0, then clearly g(c) \ge 0.
If 3b^3 + 2ab^2 + 4a^2b - 3a^3 \le 0, then g(c) is concave for all c \in [0,b], thus g(c) \ge \min\{g(0),g(b)\}.
But g(0) = ab(a - b)^4 \ge 0, and
g(b) = b \left[ \frac {1}{4}(a - b)[(2a^2 - 6ab - b^2)^2 + 43b^4] + 8b^5\right] \ge 0.
Thus, our claim is proved.
Now, back to our problem
We will consider 2 cases
Case 1. If \frac {a^2 + b^2}{c^2 + ab} \le 2, then we have
\sqrt [3]{\frac {c^2 + ab}{a^2 + b^2}} \ge \frac {1}{\sqrt [3]{2}},
and
\sqrt [3]{\frac {a^2 + bc}{b^2 + c^2}} + \sqrt [3]{\frac {b^2 + ca}{c^2 + a^2}} \ge 2,
\Rightarrow \;\; \text{done}
Case 2. If \frac {a^2 + b^2}{c^2 + ab} \ge 2, then we have
\sqrt [3]{\frac {a^2 + bc}{b^2 + c^2}} + \sqrt [3]{\frac {b^2 + ca}{c^2 + a^2}} \ge \sqrt [3]{\frac {4(a^2 + b^2)}{c^2 + ab}} = 4\sqrt [3]{4}x,
Hence
LHS \ge \sqrt [3]{4} x + \frac {1}{x} \ge 2 + \frac {1}{\sqrt [3]{2}}
(since x \ge \sqrt [3]{2})
We have done. Equality holds if and only if (a,b,c)\sim (1,1,0). Smile
Sửa bởi can_hang2007 vào lúc 26-03-2009 14:06
The love makes us stronger!
 
daogiauvang
 Problem 14(small) Let a,b,c be positive real numbers such that abc=1
Prove that:
\frac {1}{\sqrt {2a + 2ab + 1}} + \frac {1}{\sqrt {2b + 2bc + 1}} + \frac {1}{\sqrt {2c + 2ca + 1}}\geq \ 1}

Link: http://www.mathli...p?t=262429
Solution 1: (arqady)
We have (x + 1)^2 - (2x + 1) = x^2 > 0, which shows that
\frac {1}{\sqrt {2x + 1}} \ge \frac {1}{x + 1}.
Therefor, \frac {1}{\sqrt {2a + 2ab + 1}} + \frac {1}{\sqrt {2b + 2bc + 1}} + \frac {1}{\sqrt {2c + 2ca + 1}}\geq \ 1}
\sum_{cyc}\frac {1}{\sqrt {2a + 2ab + 1}}\geq\sum_{cyc}\frac {1}{a + ab + 1} = 1.

Solution 2: (can_hang)
a = \frac {x}{y},b = \frac {z}{x},c = \frac {y}{z}, then our inequality becomes
\sqrt {\frac {x}{x + 2y + 2z}} + \sqrt {\frac {y}{y + 2z + 2x}} + \sqrt {\frac {z}{z + 2x + 2y}} \ge 1,
which is trivial as
\sqrt {\frac {x}{x + 2y + 2z}} = \frac {x}{\sqrt {x(x + 2y + 2z)}} \ge \frac {2x}{x + (x + 2y + 2z)} = \frac {x}{x + y + z},
Sửa bởi daogiauvang vào lúc 26-03-2009 15:47
If I feel unhappy , I do mathematics to become happy , If I feel happy , I do mathematics to keep happy

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trungtin
 I like this topic .:B it 's very interesting
problem 15 Let a,b,c \geq 0 and a+b+c=1
prove that: \sqrt{13} \leq \sqrt{a^2+a+1}+\sqrt{b^2+b+1}+\sqrt{c^2+c+1} \leq 2+\sqrt{3}
proof
(i) Using Minkowski 's inquality
\sum{\sqrt{a^2+a+1}}=\sum{\sqrt{(a+\frac{1}{2})^2+\frac{3}{4}}} \geq \sqrt{(\sum{a}+\frac{3}{2})^2+(\frac{3\sqrt{3}}{2})^2}=\sqrt{13}
Equality hold if and if a=b=c=\frac{1}{3}
(ii) We have \sqrt{a^2+a+1}+\sqrt{b^2+b+1} \leq 1+\sqrt{(a+b)^2+(a+b)+1}
\Leftrightarrow ab(1+\sqrt{(a+b)^2+(a+b)+1}-(a+b)) \geq (true,too)
thus \sum{\sqrt{c^2+c+1}} \leq 1+\sqrt{c^2-3c+3}+\sqrt{c^2+c+1}
Let f(c)=\sqrt{c^2-3c+3}+\sqrt{c^2+c+1} since c \subset (0,1]
then maxf(c)_{(0,1]}=f(1)=1+\sqrt{3}
So \sum{\sqrt{a^2+a+1}} \leq 2+\sqrt{3}
Equality hold if and if a=b=0,c=1
Sửa bởi trungtin vào lúc 27-03-2009 06:00
tin
 
daogiauvang
 Problem 16:( Algadin)
Let a,b,c > 0 and a^3 + b^3 + c^3 = 3
Prove that

\frac {a^2}{3 - a^2} + \frac {b^2}{3 - b^2} + \frac {c^2}{3 - c^2}\geq\frac {3}{2}
Link: http://www.mathli...p?t=264617
Proof:
Using AM-GM inequality we have
\frac {{a^2 }}{{3 - a^2 }} = \frac {{a^3 }}{{a(3 - a^2 )}} = \frac {{a^3 }}{{\sqrt {a^2 (3 - a^2 )^2 } }} = \frac {{a^3 }}{{2.\sqrt {a^2 .\frac {{3 - a^2 }}{2}.\frac {{3 - a^2 }}{2}} }} \ge \frac {{a^3 }}{2}

Hence, LHS \geq \frac{3}{2}
Sửa bởi daogiauvang vào lúc 27-03-2009 13:38
If I feel unhappy , I do mathematics to become happy , If I feel happy , I do mathematics to keep happy

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daogiauvang
 Problem 17: Brazilian Math Olympiad 2008, Problem 3
Let x,y,z real numbers such that x + y + z = xy + yz + zx. Find the minimum value of
P={x \over x^2 + 1} + {y\over y^2 + 1} + {z\over z^2 + 1}

Link: http://www.mathlinks.ro/viewtopic.php?t=235748

Proof: Can_hang2007
We need to prove that

\frac {x}{x^2 + 1} + \frac {y}{y^2 + 1} + \frac {z}{z^2 + 1} \ge - \frac {1}{2}


\Leftrightarrow \frac {(x + 1)^2}{x^2 + 1} + \frac {(y + 1)^2}{y^2 + 1} \ge \frac {(z - 1)^2}{z^2 + 1}

From the given condition, we have that

z(x + y - 1) = x + y - xy

Notice that we can't have x + y = 1. Otherwise, if x + y = 1, then x + y - xy = 0 or xy = 1, which is contradition since xy \le \frac {(x + y)^2}{4} = \frac {1}{4}, therefore x + y \neq 1, hence

z = \frac {x + y - xy}{x + y - 1}
Then, we can easily rewrite our inequality as

\frac {(x + 1)^2}{x^2 + 1} + \frac {(y + 1)^2}{y^2 + 1} \ge \frac {(xy - 1)^2}{(x + y - 1)^2 + (x + y - xy)^2}
Now, using Cauchy Schwarz Inequality, we have

\frac {(x + 1)^2}{x^2 + 1} + \frac {(y + 1)^2}{y^2 + 1} \ge \frac {[(1 + x)(1 - y) + (1 + y)(1 - x)]^2}{(1 + x^2)(1 - y)^2 + (1 + y^2)(1 - x)^2} = \frac {4(xy - 1)^2}{(1 + x^2)(1 - y)^2 + (1 + y^2)(1 - x)^2}
Hence, it suffices to prove that

4(x + y - 1)^2 + 4(x + y - xy)^2 \ge (1 + x^2)(1 - y)^2 + (1 + y^2)(1 - x)^2


\Leftrightarrow f(x) = (y^2 - 3y + 3)x^2 - (3y^2 - 8y + 3)x + 3y^2 - 3y + 1 \ge 0
Now, we have

\Delta_{f} = (3y^2 - 8y + 3)^2 - 4(y^2 - 3y + 3)(3y^2 - 3y + 1) = - 3(y^2 - 1)^2 \le 0
Thus f(x) \ge 0 and we have done!
If I feel unhappy , I do mathematics to become happy , If I feel happy , I do mathematics to keep happy

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daogiauvang
 Problem 17: (antiparticle)
Prove that: \ 3(a^3 + b^3 + c^3 + abc) \ge 4(a^2b + b^2c + c^2a)

for all positive real numbers a, b, c.

Link: http://www.mathli...p?t=256801
Proof: (NextPeace)

3(a^3 + b^3 + c^3 + abc) \ge 2(a^3 + b^3 + c^3) + (a^2b + b^2c + c^2a) + (ab^2 + bc^2 + ca^2)\\ = \sum (a^3 + ab^2) + \sum a^2b + \frac {\sum(a^3 + a^3 + b^3)}{3}\ge 2\sum a^2b + \sum a^2b + \sum a^2b \\ = 4(a^2b + b^2c + c^2a)
Sửa bởi daogiauvang vào lúc 27-03-2009 14:08
If I feel unhappy , I do mathematics to become happy , If I feel happy , I do mathematics to keep happy

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trungtin
 Problem 18(ductrung)
For any positive real numbers and , prove that \sqrt[6]{\frac{a+b}{a+c}}+\sqrt[6]{\frac{b+c}{b+a}}+\sqrt[6]{\frac{c+a}{c+b}}\le \frac{a+b+c}{\sqrt[3]{abc}}
link: http://www.mathli...p?t=111171
Proof (lovasz)
\left(\sqrt[3]{\frac{a+b}{a+c}}+\sqrt[3]{\frac{b+c}{b+a}}+\sqrt[3]{\frac{c+a}{c+b}}\right)^{3}\\ \le 6(a+b+c)\left({1 \over a+b}+{1 \over b+c}+{1 \over c+a}\right) \\ \le{3(a+b+c)\left({1 \over a}+{1 \over b}+{1 \over c}\right).
The rest is:

{(a+b+c)^{3}\over abc}\ge{3(a+b+c)}\left({1 \over a}+{1 \over b}+{1 \over c}\right) \\ \iff (a+b+c)^{2}\ge 3(ab+bc+ca).
Proof complete.
Sửa bởi trungtin vào lúc 27-03-2009 16:53
tin
 
daogiauvang
 Problem 19:(nguoivn)
Given a, b, c \geq\ 0 satisfy a + b + c = 3. Prove that: \sqrt {a + bc} + \sqrt {b + ca} + \sqrt {c + ab}\geq \ \sqrt {2}(ab + bc + ca)

Link: http://www.mathli...p?t=268894
Proof:

Solution 1: ( arqady)

By Holder we obtain:
\left(\sum_{cyc}\sqrt {a + bc}\right)^2\sum_{cyc}\frac {1}{a + bc}\geq27.
Thus, we need to prove that 27\geq2(ab + ac + bc)^2\sum_{cyc}\frac {1}{a + bc}, which is obvious.


Solution 2(ElChapin):

By AMGM we have that
\sum{\sqrt {a + bc} + \sqrt {a + bc} + \dfrac{(a + bc)^2}{2\sqrt {2}}}\ge\sum{\dfrac{3(a + bc)}{\sqrt {2}}}

So it suffices to prove that
\sum{\dfrac{3(a + bc)}{\sqrt {2}} - \dfrac{(a + bc)^2}{2}}\le 2\sqrt {2}(ab + bc + ca)

Which is true because we have that
(a + b + c)^2\ge 3(ab + bc + ca)\Leftrightarrow a + b + c\ge ab + bc + ca\Leftrightarrow

2(a + b + c)^2\ge (ab + bc + ca)^2 + (a + b + c)^2\Leftrightarrow

6(a + b + c)\ge \sum{b^2c^2 + 2abc + a^2 + 2bc}\Leftrightarrow

6\sum{(a + bc)}\ge \sum{(a + bc)^2 + 8bc}\Leftrightarrow

\sum{\dfrac{3(a + bc)}{\sqrt {2}} - \dfrac{(a + bc)^2}{2}}\le 2\sqrt {2}(ab + bc + ca)
If I feel unhappy , I do mathematics to become happy , If I feel happy , I do mathematics to keep happy

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daogiauvang
 Problem 20: (nguoivn) Given a, b, c > 0. Prove that: a + b + c + \frac {1}{abc} \geq\ \frac {2(ab + bc + ca + 1)^2}{(a + b)(b + c)(c + a)}
Link: http://www.mathli...p?t=278376
Proof: ( can_hang2007)
Applying the Cauchy Schwarz Inequality, we have
LHS\cdot (ab^2 + bc^2 + ca^2 + abc) \ge (ab + bc + ca + 1)^2,
and
LHS \cdot (ac^2 + ba^2 + cb^2 + abc) \ge (ca + ab + bc + 1)^2.
It follows that
LHS \ge \frac {(ab + bc + ca + 1)^2}{ab^2 + bc^2 + ca^2 + abc},
and
LHS \ge \frac {(ab + bc + ca + 1)^2}{a^2b + b^2c + c^2a + abc}.
Therefore
2LHS \ge (ab + bc + ca + 1)^2\left( \frac {1}{a^2b + b^2c + c^2a + abc} + \frac {1}{ab^2 + bc^2 + ca^2 + abc}\right) \ge \frac {4 (ab + bc + ca + 1)^2}{(a^2b + b^2c + c^2a + abc) + (ab^2 + bc^2 + ca^2 + abc)} = \frac {4(ab + bc + ca + 1)^2}{(a + b)(b + c)(c + a)}.
So, by dividing both sides for 2, we obtain the result.

If I feel unhappy , I do mathematics to become happy , If I feel happy , I do mathematics to keep happy

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18/08/2013 05:31
Diễn đàn mình nhiều bài hay và chất lượng quá. Em mong diễn đàn ta cứ tồn tại mãi để chúng em còn được tiếp cận với các tài liệu do các anh viết. Smile

16/08/2013 17:50
Nhưng các bài chất lượng thì vẫn còn đây!

25/07/2013 16:51
Sad diễn đàn ít có hoạt động nhỉ?

23/07/2013 07:06
Kvant, Vualangbat, Hoa dai, Nguyen Ngoc...

20/07/2013 08:20
Các Admin có những ai anh nhỉ ??

18/07/2013 20:26
e cứ đợi các admins tụ tập lại 1 lần thảo luận đã, giờ admins trốn hết rồi

11/07/2013 07:16
Bây giờ làm thế nào để diễn đàn được như trước nhỉ ??

02/06/2013 08:20
nhưng chưa có chiều sâu, vì các admin chủ lực đang bận bịu gì đó và ko có liên lạc lẫn nhau.

31/05/2013 07:00
Phải nói là trong số các diễn đàn toán thì em thấy diễn đàn ta là đẹp nhất. Wink

04/03/2013 14:16
thi Toán đơn giản mà. E cần dịch gì a dịch cho, qui đổi theo bài theo thời gian khoảng 2-3 tháng e đọc hiểu và đóan vô tư.

02/03/2013 19:10
Thuê thế nào anh ?? Grin

22/02/2013 13:11
Can thue nguoi ko a day cho) Khoang 3 thang la doc dich duoc

22/02/2013 06:47
Nhìn mà thèm học Tiếng Nga Smile

05/02/2013 20:05
Quet' nha chuan bi don tet

28/01/2013 06:08
Tuan Anh, sao kho' du vay la sao e?

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