Question

Proving $(a+b+c)^2\prod_{cyc}(a+b)-4\sum_{cyc}(a^2b+a^2c)\sum_{cyc}ab\geqq 0$

score 125 · Answer 1 · 2020-05-16 11:11:42

125

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Proving $(a+b+c)^2\prod_{cyc}(a+b)-4\sum_{cyc}(a^2b+a^2c)\sum_{cyc}ab\geqq 0$

Published on 16 May 2020 - 11:11

score 160 · Answer 2 · 2020-05-17 13:03:41

160

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Prove the following inequality $\sum_{i<j<k}\frac{a_ia_ja_k}{(n-2)(n-1)n}\le \bigg(\sum_{i<j}\frac{a_ia_j}{(n-1)n}\bigg)^2+\frac{1}{12}$

Published on 17 May 2020 - 13:03

score 198 · Answer 3 · 2020-05-21 00:44:21

198

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Prove $\frac{x^2+yz}{\sqrt{2x^2(y+z)}}+\frac{y^2+zx}{\sqrt{2y^2(z+x)}}+\frac{z^2+xy}{\sqrt{2z^2(x+y)}}\geqq 1$

Published on 21 May 2020 - 0:44

score 163 · Answer 4 · 2020-05-27 12:00:34

163

Views

Proving inequality by SOS.

Published on 27 May 2020 - 12:00

score 98 · Answer 5 · 2026-03-25 12:49:53

98

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Prove$:$ $\sum\limits_{cyc} (\frac{a}{b+c}-\frac{1}{2}) \geqq (\sum\limits_{cyc} ab)\Big[\sum\limits_{cyc} \frac{1}{(a+b)^2}\Big]-\frac{9}{4}$

Published on 25 Mar 2026 - 12:49

score 107 · Answer 6 · 2020-06-04 15:42:37

107

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Let $a, b, c>0$. Prove that $\sum \limits_{cyc}{\frac{a}{b+c}\left(\frac{b}{c+a}+\frac{c}{a+b}\right)}\le \frac{(a+b+c)^2}{2(ab+bc+ca)}$

Published on 04 Jun 2020 - 15:42

score 104 · Answer 7 · 2020-06-05 17:07:18

104

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Let $a,$ $b$ and $c$ are positive numbers.

Published on 05 Jun 2020 - 17:07

score 72 · Answer 8 · 2020-06-07 08:47:23

72

Views

Let $a,b,c>0$ then prove this inequality holds

Published on 07 Jun 2020 - 8:47

score 185 · Answer 9 · 2020-06-17 08:03:07

185

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Given three real numbers $a,b,c$ so that $\{a, b, c\}\subset [1, 2]$ . Prove that $7abc\geq ab(a+ b)+ bc(b+ c)+ ca(c+ a)$ .

Published on 17 Jun 2020 - 8:03

score 141 · Answer 10 · 2020-07-05 13:54:58

141

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For which $k$ does $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+k-3\ge \left(2+\frac k3\right)\cdot \frac{a+b+c}{\sqrt[3]{abc}}$ hold?

Published on 05 Jul 2020 - 13:54

score 99 · Answer 11 · 2026-02-23 04:22:02

99

Views

Proving a non-homogeneous inequality with $x,y,z>0$

Published on 23 Feb 2026 - 4:22

score 147 · Answer 12 · 2020-07-11 11:30:19

147

Views

An inequality involving real numbers

Published on 11 Jul 2020 - 11:30

score 143 · Answer 13 · 2020-07-15 13:42:09

143

Views

An stronger inequality than in AoPS.

Published on 15 Jul 2020 - 13:42

score 118 · Answer 14 · 2020-07-23 13:09:27

118

Views

Prove that in a triangle $\sum\limits_{cyc}\frac{w_bw_c}{w_a}\geq\frac{3}{4}\left(\sum\limits_{cyc}\frac{a^2w_a}{w_bw_c}\right)\geq\sqrt{3}s$

Published on 23 Jul 2020 - 13:09

score 175 · Answer 15 · 2026-03-25 12:49:57

175

Views

Proving $\frac {a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} \geqslant \frac 32 \cdot \sqrt[6]{\frac{ab+bc+ca}{a^2+b^2+c^2}}$

Published on 25 Mar 2026 - 12:49

List Question

Proving $(a+b+c)^2\prod_{cyc}(a+b)-4\sum_{cyc}(a^2b+a^2c)\sum_{cyc}ab\geqq 0$

Prove the following inequality $\sum_{i<j<k}\frac{a_ia_ja_k}{(n-2)(n-1)n}\le \bigg(\sum_{i<j}\frac{a_ia_j}{(n-1)n}\bigg)^2+\frac{1}{12}$

Prove $\frac{x^2+yz}{\sqrt{2x^2(y+z)}}+\frac{y^2+zx}{\sqrt{2y^2(z+x)}}+\frac{z^2+xy}{\sqrt{2z^2(x+y)}}\geqq 1$

Proving inequality by SOS.

Prove$:$ $\sum\limits_{cyc} (\frac{a}{b+c}-\frac{1}{2}) \geqq (\sum\limits_{cyc} ab)\Big[\sum\limits_{cyc} \frac{1}{(a+b)^2}\Big]-\frac{9}{4}$

Let $a, b, c>0$. Prove that $\sum \limits_{cyc}{\frac{a}{b+c}\left(\frac{b}{c+a}+\frac{c}{a+b}\right)}\le \frac{(a+b+c)^2}{2(ab+bc+ca)}$

Let $a,$ $b$ and $c$ are positive numbers.

Let $a,b,c>0$ then prove this inequality holds

Given three real numbers $a,b,c$ so that $\{a, b, c\}\subset [1, 2]$ . Prove that $7abc\geq ab(a+ b)+ bc(b+ c)+ ca(c+ a)$ .

For which $k$ does $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+k-3\ge \left(2+\frac k3\right)\cdot \frac{a+b+c}{\sqrt[3]{abc}}$ hold?

Proving a non-homogeneous inequality with $x,y,z>0$

An inequality involving real numbers

An stronger inequality than in AoPS.

Prove that in a triangle $\sum\limits_{cyc}\frac{w_bw_c}{w_a}\geq\frac{3}{4}\left(\sum\limits_{cyc}\frac{a^2w_a}{w_bw_c}\right)\geq\sqrt{3}s$

Proving $\frac {a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} \geqslant \frac 32 \cdot \sqrt[6]{\frac{ab+bc+ca}{a^2+b^2+c^2}}$

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