Question

Prove the inequality $\sum_{cyc}\frac{a^3}{b\sqrt{a^3+8}}\ge 1$

score 98 · Answer 1 · 2020-07-29 05:32:03

98

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Prove the inequality $\sum_{cyc}\frac{a^3}{b\sqrt{a^3+8}}\ge 1$

Published on 29 Jul 2020 - 5:32

score 141 · Answer 2 · 2020-07-30 16:18:11

141

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Show that $(a^3+a+1)(b^3+b+1)(c^3+c+1)\le 27$

Published on 30 Jul 2020 - 16:18

score 839 · Answer 3 · 2020-07-30 18:09:28

839

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If $a$, $b$, $c$, $d$ are positive reals so $(a+c)(b+d) = 1$, prove the following inequality would be greater than or equal to $\frac {1}{3}$.

Published on 30 Jul 2020 - 18:09

score 133 · Answer 4 · 2016-07-30 10:05:37

133

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Proving that $\frac{\sqrt{1-x^2_{1}}+\sqrt{1-x^2_{2}}+\ldots+\sqrt{1-x^2_{17}}}{x_{1}+x_{2}+\ldots+x_{17}}\geq 4$

Published on 30 Jul 2016 - 10:05

score 76 · Answer 5 · 2016-08-15 16:53:44

76

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Inequality on non-negative reals

Published on 15 Aug 2016 - 16:53

score 306 · Answer 6 · 2016-09-05 04:25:44

306

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Inequality with $a+b+c=1$

Published on 05 Sep 2016 - 4:25

score 113 · Answer 7 · 2016-09-10 19:48:33

113

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${\sqrt{a+b+c} + \sqrt{a}\over b+c} + {\sqrt{a+b+c} + \sqrt{b}\over a+c} +{\sqrt{a+b+c}+ \sqrt{c}\over b+a} \ge {9-3\sqrt{3}\over2\sqrt{a+b+c}}$

Published on 10 Sep 2016 - 19:48

score 209 · Answer 8 · 2016-11-19 14:43:22

209

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Prove that $\triangle ABC$ is equilateral, If $(a+b-c)^{a}\cdot (b+c-a)^{b}\cdot (c+a-b)^{c}\geq a^a\cdot b^b\cdot c^c$

Published on 19 Nov 2016 - 14:43

score 221 · Answer 9 · 2016-12-23 03:23:30

221

Views

prove $(a^3+1)(b^3+1)(c^3+1)\ge 8$

Published on 23 Dec 2016 - 3:23

score 192 · Answer 10 · 2016-12-29 17:46:51

192

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How to prove$\sum\frac{a}{a^2+a+1}\leq\sum\frac{1}{a^2+a+1}$

Published on 29 Dec 2016 - 17:46

score 305 · Answer 11 · 2017-01-03 10:31:45

305

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Given that $xyz=1$, prove that $\frac{x}{1+x^4}+\frac{y}{1+y^4}+\frac{z}{1+z^4}\le \frac{3}{2}$

Published on 03 Jan 2017 - 10:31

score 738 · Answer 12 · 2017-01-12 17:54:56

738

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Proving $\left(\frac{a}{a+b+c}\right)^2+\left(\frac{b}{b+c+d}\right)^2+\left(\frac{c}{c+d+a}\right)^2+\left(\frac{d}{d+a+b}\right)^2\ge\frac{4}{9}$

Published on 12 Jan 2017 - 17:54

score 349 · Answer 13 · 2026-03-23 10:24:54

349

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Proof that $\frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b} \ge \frac{3}{7}$ for all $a,b,c > 0$

Published on 23 Mar 2026 - 10:24

score 221 · Answer 14 · 2017-01-29 14:31:16

221

Views

Prove that $\sum\limits_{cyc} a^7 \geq \sum\limits_{cyc}a^4b^3$

Published on 29 Jan 2017 - 14:31

score 118 · Answer 15 · 2017-03-04 11:22:24

118

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Prove that $\frac{(2a+b+c)^2}{2a^2+(b+c)^2}+\frac{(2b+c+a)^2}{2b^2+(c+a)^2}+\frac{(2c+a+b)^2}{2c^2+(a+b)^2} \le 8$

Published on 04 Mar 2017 - 11:22

List Question

Prove the inequality $\sum_{cyc}\frac{a^3}{b\sqrt{a^3+8}}\ge 1$

Show that $(a^3+a+1)(b^3+b+1)(c^3+c+1)\le 27$

If $a$, $b$, $c$, $d$ are positive reals so $(a+c)(b+d) = 1$, prove the following inequality would be greater than or equal to $\frac {1}{3}$.

Proving that $\frac{\sqrt{1-x^2_{1}}+\sqrt{1-x^2_{2}}+\ldots+\sqrt{1-x^2_{17}}}{x_{1}+x_{2}+\ldots+x_{17}}\geq 4$

Inequality on non-negative reals

Inequality with $a+b+c=1$

${\sqrt{a+b+c} + \sqrt{a}\over b+c} + {\sqrt{a+b+c} + \sqrt{b}\over a+c} +{\sqrt{a+b+c}+ \sqrt{c}\over b+a} \ge {9-3\sqrt{3}\over2\sqrt{a+b+c}}$

Prove that $\triangle ABC$ is equilateral, If $(a+b-c)^{a}\cdot (b+c-a)^{b}\cdot (c+a-b)^{c}\geq a^a\cdot b^b\cdot c^c$

prove $(a^3+1)(b^3+1)(c^3+1)\ge 8$

How to prove$\sum\frac{a}{a^2+a+1}\leq\sum\frac{1}{a^2+a+1}$

Given that $xyz=1$, prove that $\frac{x}{1+x^4}+\frac{y}{1+y^4}+\frac{z}{1+z^4}\le \frac{3}{2}$

Proving $\left(\frac{a}{a+b+c}\right)^2+\left(\frac{b}{b+c+d}\right)^2+\left(\frac{c}{c+d+a}\right)^2+\left(\frac{d}{d+a+b}\right)^2\ge\frac{4}{9}$

Proof that $\frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b} \ge \frac{3}{7}$ for all $a,b,c > 0$

Prove that $\sum\limits_{cyc} a^7 \geq \sum\limits_{cyc}a^4b^3$

Prove that $\frac{(2a+b+c)^2}{2a^2+(b+c)^2}+\frac{(2b+c+a)^2}{2b^2+(c+a)^2}+\frac{(2c+a+b)^2}{2c^2+(a+b)^2} \le 8$

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