Question

Bounds on the AM-GM gap?

score 94 · Answer 1 · 2021-02-15 11:15:47

94

Views

Bounds on the AM-GM gap?

Published on 15 Feb 2021 - 11:15

score 61 · Answer 2 · 2021-02-15 13:03:00

61

Views

Notation difficulty in AM-GM inequality proof

Published on 15 Feb 2021 - 13:03

score 340 · Answer 3 · 2021-02-16 14:36:26

340

Views

Using AM-GM to find the minimum

Published on 16 Feb 2021 - 14:36

score 142 · Answer 4 · 2021-02-20 02:53:24

142

Views

Prove that $2^{n}>1+n\sqrt{2^{\left(n-1\right)}}, \ \ \forall n>2$.

Published on 20 Feb 2021 - 2:53

score 200 · Answer 5 · 2021-02-21 07:28:47

200

Views

Prove $ \sum \sqrt{\tan A} \geq \sum \sqrt{\cot\frac{A}{2}}$

Published on 21 Feb 2021 - 7:28

score 100 · Answer 6 · 2021-02-22 13:05:25

100

Views

Prove $\frac1{\sin A}+\frac1{\sin B}+\frac1{\sin C}-\frac12(\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}) \ge \sqrt{3}$

Published on 22 Feb 2021 - 13:05

score 104 · Answer 7 · 2021-02-23 07:18:44

104

Views

Better way to prove $\frac{1}{4a^2+3b^2+2c^2}\leq |\sqrt[3]{4}\, a+\sqrt[3]{2}\,b+c|$

Published on 23 Feb 2021 - 7:18

score 100 · Answer 8 · 2026-03-27 06:01:40

100

Views

$\sum\sqrt{\frac{2a}{b+c}}\le\sqrt[3]{9\sum\frac{a}{b}}$

Published on 27 Mar 2026 - 6:01

score 81 · Answer 9 · 2021-02-25 12:03:46

81

Views

Question regarding a proof of a inequality

Published on 25 Feb 2021 - 12:03

score 175 · Answer 10 · 2021-02-26 02:36:14

175

Views

How can we prove the following product-sum inequality involving e?

Published on 26 Feb 2021 - 2:36

score 243 · Answer 11 · 2021-02-26 05:44:37

243

Views

Supremum of $x^2y^2(x^2+y^2)$

Published on 26 Feb 2021 - 5:44

score 110 · Answer 12 · 2026-03-27 01:04:16

110

Views

Question about Sui Zhen Lin's proof for inequality $\sqrt{\frac{a^2}{9b^2-8b+4}}+\sqrt{\frac{4b}{a+4}}\leq 1$ with positive numbers $a,b$ so $a+b=1$

Published on 27 Mar 2026 - 1:04

score 58 · Answer 13 · 2021-02-27 03:59:48

58

Views

Minimize $z=e^{-x_{1}}+e^{-2 x_{2}}$ Subject to $\quad x_{1}+x_{2} \leq 1 \ , x_{1}, x_{2} \geq 0$ without KKT

Published on 27 Feb 2021 - 3:59

score 88 · Answer 14 · 2026-02-23 18:59:46

88

Views

Given 4 numbers $a, b, c, d> 0,$ show $16\max\limits_{\bigcirc}\left \{ a^{3}+ 3bcd \right \}\!\geq\!\left ( a+ b+ c+ d \right )^{3}$

Published on 23 Feb 2026 - 18:59

score 457 · Answer 15 · 2021-03-03 19:15:14

457

Views

If $x,y,z>0$, prove that: $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \sqrt{2}\sqrt{2-\frac{7xyz}{(x+y)(y+z)(x+z)}}$

Published on 03 Mar 2021 - 19:15

List Question

Bounds on the AM-GM gap?

Notation difficulty in AM-GM inequality proof

Using AM-GM to find the minimum

Prove that $2^{n}>1+n\sqrt{2^{\left(n-1\right)}}, \ \ \forall n>2$.

Prove $ \sum \sqrt{\tan A} \geq \sum \sqrt{\cot\frac{A}{2}}$

Prove $\frac1{\sin A}+\frac1{\sin B}+\frac1{\sin C}-\frac12(\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}) \ge \sqrt{3}$

Better way to prove $\frac{1}{4a^2+3b^2+2c^2}\leq |\sqrt[3]{4}\, a+\sqrt[3]{2}\,b+c|$

$\sum\sqrt{\frac{2a}{b+c}}\le\sqrt[3]{9\sum\frac{a}{b}}$

Question regarding a proof of a inequality

How can we prove the following product-sum inequality involving e?

Supremum of $x^2y^2(x^2+y^2)$

Question about Sui Zhen Lin's proof for inequality $\sqrt{\frac{a^2}{9b^2-8b+4}}+\sqrt{\frac{4b}{a+4}}\leq 1$ with positive numbers $a,b$ so $a+b=1$

Minimize $z=e^{-x_{1}}+e^{-2 x_{2}}$ Subject to $\quad x_{1}+x_{2} \leq 1 \ , x_{1}, x_{2} \geq 0$ without KKT

Given 4 numbers $a, b, c, d> 0,$ show $16\max\limits_{\bigcirc}\left \{ a^{3}+ 3bcd \right \}\!\geq\!\left ( a+ b+ c+ d \right )^{3}$

If $x,y,z>0$, prove that: $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \sqrt{2}\sqrt{2-\frac{7xyz}{(x+y)(y+z)(x+z)}}$

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