Question

$a_1,a_2,a_3,b_1,b_2,b_3$ are positive real numbers, show: $\sqrt[3]{(a_1+b_1)(a_2+b_2)(a_3+b_3)} \geq \sqrt[3]{a_1a_2a_3} + \sqrt[3]{b_1b_2b_3}$

score 161 · Answer 1 · 2015-07-14 14:43:48

161

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$a_1,a_2,a_3,b_1,b_2,b_3$ are positive real numbers, show: $\sqrt[3]{(a_1+b_1)(a_2+b_2)(a_3+b_3)} \geq \sqrt[3]{a_1a_2a_3} + \sqrt[3]{b_1b_2b_3}$

Published on 14 Jul 2015 - 14:43

score 748 · Answer 2 · 2015-07-22 01:53:26

748

Views

Prove the inequality $\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$ with the constraint $abc=1$

Published on 22 Jul 2015 - 1:53

score 925 · Answer 3 · 2015-07-22 14:22:22

925

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$\sum\limits_{i=1}^n \frac{x_i}{\sqrt[n]{x_i^n+(n^n-1)\prod \limits_{j=1}^nx_j}} \ge 1$, for all $x_i>0.$

Published on 22 Jul 2015 - 14:22

score 105 · Answer 4 · 2015-10-24 09:38:42

105

Views

Dospinescu's Inequality

Published on 24 Oct 2015 - 9:38

score 2.1k · Answer 5 · 2016-01-20 03:32:52

2.1k

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Prove $\sum\limits_{\mathrm{cyc}} \sqrt{a^3+2} \ge \sqrt{\frac{9+3\sqrt{3}}{2}(a^2+b^2+c^2)}$ for $a, b, c \ge 0$

Published on 20 Jan 2016 - 3:32

score 1.9k · Answer 6 · 2016-05-09 07:16:38

1.9k

Views

Discreet weighted mean inequality

Published on 09 May 2016 - 7:16

score 112 · Answer 7 · 2013-08-20 04:08:03

112

Views

A $4$ variable inequality

Published on 20 Aug 2013 - 4:08

score 480 · Answer 8 · 2014-02-16 17:00:47

480

Views

How to prove this inequality $\frac{a}{\sqrt{b^2+b+1}}+\frac{b}{\sqrt{c^2+c+1}}+\frac{c}{\sqrt{a^2+a+1}}\ge\sqrt{3}\;?$

Published on 16 Feb 2014 - 17:00

score 417 · Answer 9 · 2014-02-17 14:37:58

417

Views

Proof of one inequality $a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$

Published on 17 Feb 2014 - 14:37

score 201 · Answer 10 · 2014-03-04 17:26:56

201

Views

If $a_1a_2\cdots a_n =1$, then $\prod_{i=1}^n (1+a_i) \geq 2^n$

Published on 04 Mar 2014 - 17:26

score 330 · Answer 11 · 2017-04-13 19:45:51

330

Views

Show $(1+x_1)(1+x_2)...(1+x_n)\geq2^n$ given $x_1x_2...x_n=1$

Published on 13 Apr 2017 - 19:45

score 508 · Answer 12 · 2017-04-20 08:02:24

508

Views

Finding maximum value of $x^2y$ given following condition.

Published on 20 Apr 2017 - 8:02

score 246 · Answer 13 · 2026-03-25 12:27:07

246

Views

$a+b+c = 3$, prove that :$a\sqrt{a+3}+b\sqrt{b+3}+c\sqrt{c+3} \geq 6$

Published on 25 Mar 2026 - 12:27

score 201 · Answer 14 · 2017-05-03 13:57:58

201

Views

Prove an inequality using Jensen's

Published on 03 May 2017 - 13:57

score 239 · Answer 15 · 2017-05-03 14:15:21

239

Views

Inequality $\frac{b}{ab+b+1} + \frac{c}{bc+c+1} + \frac{a}{ac+a+1} \ge \frac{3m}{m^2+m+1}$

Published on 03 May 2017 - 14:15

List Question

$a_1,a_2,a_3,b_1,b_2,b_3$ are positive real numbers, show: $\sqrt[3]{(a_1+b_1)(a_2+b_2)(a_3+b_3)} \geq \sqrt[3]{a_1a_2a_3} + \sqrt[3]{b_1b_2b_3}$

Prove the inequality $\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$ with the constraint $abc=1$

$\sum\limits_{i=1}^n \frac{x_i}{\sqrt[n]{x_i^n+(n^n-1)\prod \limits_{j=1}^nx_j}} \ge 1$, for all $x_i>0.$

Dospinescu's Inequality

Prove $\sum\limits_{\mathrm{cyc}} \sqrt{a^3+2} \ge \sqrt{\frac{9+3\sqrt{3}}{2}(a^2+b^2+c^2)}$ for $a, b, c \ge 0$

Discreet weighted mean inequality

A $4$ variable inequality

How to prove this inequality $\frac{a}{\sqrt{b^2+b+1}}+\frac{b}{\sqrt{c^2+c+1}}+\frac{c}{\sqrt{a^2+a+1}}\ge\sqrt{3}\;?$

Proof of one inequality $a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$

If $a_1a_2\cdots a_n =1$, then $\prod_{i=1}^n (1+a_i) \geq 2^n$

Show $(1+x_1)(1+x_2)...(1+x_n)\geq2^n$ given $x_1x_2...x_n=1$

Finding maximum value of $x^2y$ given following condition.

$a+b+c = 3$, prove that :$a\sqrt{a+3}+b\sqrt{b+3}+c\sqrt{c+3} \geq 6$

Prove an inequality using Jensen's

Inequality $\frac{b}{ab+b+1} + \frac{c}{bc+c+1} + \frac{a}{ac+a+1} \ge \frac{3m}{m^2+m+1}$

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