Question

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

score 99 · Answer 1 · 2026-03-25 11:03:42

99

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

Published on 25 Mar 2026 - 11:03

score 142 · Answer 2 · 2026-03-25 14:23:34

142

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

Published on 25 Mar 2026 - 14:23

score 840 · Answer 3 · 2026-03-25 09:29:57

840

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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 25 Mar 2026 - 9:29

score 136 · Answer 4 · 2026-04-17 12:21:21

136

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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 17 Apr 2026 - 12:21

score 79 · Answer 5 · 2026-04-14 23:17:13

79

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Inequality on non-negative reals

Published on 14 Apr 2026 - 23:17

score 309 · Answer 6 · 2026-04-30 09:05:10

309

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Inequality with $a+b+c=1$

Published on 30 Apr 2026 - 9:05

score 116 · Answer 7 · 2026-04-17 09:05:52

116

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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 17 Apr 2026 - 9:05

score 212 · Answer 8 · 2026-04-16 13:20:39

212

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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 16 Apr 2026 - 13:20

score 224 · Answer 9 · 2026-04-17 09:45:15

224

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

Published on 17 Apr 2026 - 9:45

score 195 · Answer 10 · 2026-04-13 21:08:11

195

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

Published on 13 Apr 2026 - 21:08

score 308 · Answer 11 · 2026-04-17 08:02:22

308

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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 17 Apr 2026 - 8:02

score 739 · Answer 12 · 2026-03-25 09:29:06

739

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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 25 Mar 2026 - 9:29

score 350 · Answer 13 · 2026-03-25 09:29:01

350

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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 25 Mar 2026 - 9:29

score 222 · Answer 14 · 2026-03-25 09:26:56

222

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

Published on 25 Mar 2026 - 9:26

score 121 · Answer 15 · 2026-04-16 16:15:14

121

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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 16 Apr 2026 - 16:15

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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