Question

Prove that inequality $\sum_{cyc} \frac{a}{b+2}\le 1$

score 175 · Answer 1 · 2017-07-12 17:48:39

175

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Prove that inequality $\sum_{cyc} \frac{a}{b+2}\le 1$

Published on 12 Jul 2017 - 17:48

score 80 · Answer 2 · 2017-07-14 16:38:33

80

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let $a,b,c \in \mathbb{R^+} \ \ a+b+c =1$ Then prove that : $a^3+b^3+c^3 \geq \dfrac{1}{3}(a^2+b^2+c^2)$

Published on 14 Jul 2017 - 16:38

score 632 · Answer 3 · 2017-07-22 15:06:37

632

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Rearrangement Inequality Theorem - Proof

Published on 22 Jul 2017 - 15:06

score 112 · Answer 4 · 2017-08-05 16:37:16

112

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Prove this inequality $\sum_{cyc}\frac{ab}{c\left(c+a\right)}\ge \sum_{cyc}\frac{a}{a+c}$

Published on 05 Aug 2017 - 16:37

score 102 · Answer 5 · 2017-08-13 07:07:34

If $0\le x_1\le\dots\le x_n\le1$, then $\sum\limits_{i=1}^n\left(x_i-\frac{i}{n+1}\right)^2\le\sum\limits_{i=1}^n\left(\frac{i}{n+1}\right)^2$

score 111 · Answer 6 · 2026-03-25 13:51:47

111

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Chebyshev Inequality toughnut

Published on 25 Mar 2026 - 13:51

score 201 · Answer 7 · 2017-08-15 11:55:17

201

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Inequality : $\sum_{cyc}\frac{\sqrt{a^3c}}{2\sqrt{b^3a}+3bc}\geq \frac{3}{5}$

Published on 15 Aug 2017 - 11:55

score 524 · Answer 8 · 2017-08-18 14:30:19

524

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How to prove specific inequality, assuming $\prod\limits_{i=1}^{n}(a_{i}-1)=1$

Published on 18 Aug 2017 - 14:30

score 405 · Answer 9 · 2017-08-20 07:05:14

405

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Inequality : $ (a_1a_2+a_2a_3+\ldots+a_na_1)\left(\frac{a_1}{a^2_2+a_2}+\frac{a_2}{a^2_3+a_3}+ \ldots+\frac{a_n}{a^2_1+a_1}\right)\geq \frac{n}{n+1} $

Published on 20 Aug 2017 - 7:05

score 177 · Answer 10 · 2017-08-22 13:00:07

177

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If a, b, c>0 show that $\frac{a^2}{b^2+bc}+\frac{b^2}{c^2+ac}+\frac{c^2}{a^2+ab} \ge \frac{3}{2}$

Published on 22 Aug 2017 - 13:00

score 343 · Answer 11 · 2017-08-28 14:12:24

343

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How prove this inequality $x^3y+y^3z+z^3x\ge xyz(x+y+z)$

Published on 28 Aug 2017 - 14:12

score 177 · Answer 12 · 2017-09-14 03:37:47

177

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Prove that: $\sum_{i=1}^n a_i^2\geq \sum_{i=1}^n a_ib_i$

Published on 14 Sep 2017 - 3:37

score 649 · Answer 13 · 2017-09-21 19:13:24

649

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If $a$ and $b$ are positive real numbers such that $a+b=1$, then prove that $\big(a+\frac{1}{a}\big)^2 +\big(b+\frac{1}{b}\big)^2 \ge\frac{25}{2}$

Published on 21 Sep 2017 - 19:13

score 51 · Answer 14 · 2017-09-27 09:07:06

51

Views

Smallest Sum of products, under cyclic permutations

Published on 27 Sep 2017 - 9:07

score 131 · Answer 15 · 2017-10-03 10:59:16

131

Views

Does $(\sum_{i=1}^n a_i^{1.5})^2 - \sum_{i=1}^n a_i \; \sum_{i=1}^n a_i a_{i+1} > 0$ hold for $n\le 16$?

Published on 03 Oct 2017 - 10:59

List Question

Prove that inequality $\sum_{cyc} \frac{a}{b+2}\le 1$

let $a,b,c \in \mathbb{R^+} \ \ a+b+c =1$ Then prove that : $a^3+b^3+c^3 \geq \dfrac{1}{3}(a^2+b^2+c^2)$

Rearrangement Inequality Theorem - Proof

Prove this inequality $\sum_{cyc}\frac{ab}{c\left(c+a\right)}\ge \sum_{cyc}\frac{a}{a+c}$

If $0\le x_1\le\dots\le x_n\le1$, then $\sum\limits_{i=1}^n\left(x_i-\frac{i}{n+1}\right)^2\le\sum\limits_{i=1}^n\left(\frac{i}{n+1}\right)^2$

Chebyshev Inequality toughnut

Inequality : $\sum_{cyc}\frac{\sqrt{a^3c}}{2\sqrt{b^3a}+3bc}\geq \frac{3}{5}$

How to prove specific inequality, assuming $\prod\limits_{i=1}^{n}(a_{i}-1)=1$

Inequality : $ (a_1a_2+a_2a_3+\ldots+a_na_1)\left(\frac{a_1}{a^2_2+a_2}+\frac{a_2}{a^2_3+a_3}+ \ldots+\frac{a_n}{a^2_1+a_1}\right)\geq \frac{n}{n+1} $

If a, b, c>0 show that $\frac{a^2}{b^2+bc}+\frac{b^2}{c^2+ac}+\frac{c^2}{a^2+ab} \ge \frac{3}{2}$

How prove this inequality $x^3y+y^3z+z^3x\ge xyz(x+y+z)$

Prove that: $\sum_{i=1}^n a_i^2\geq \sum_{i=1}^n a_ib_i$

If $a$ and $b$ are positive real numbers such that $a+b=1$, then prove that $\big(a+\frac{1}{a}\big)^2 +\big(b+\frac{1}{b}\big)^2 \ge\frac{25}{2}$

Smallest Sum of products, under cyclic permutations

Does $(\sum_{i=1}^n a_i^{1.5})^2 - \sum_{i=1}^n a_i \; \sum_{i=1}^n a_i a_{i+1} > 0$ hold for $n\le 16$?

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