Question

Natural numbers inequality $na^{n-1}b\leq(n-1)a^n+b^n$ by induction

score 118 · Answer 1 · 2020-02-23 18:50:19

118

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Natural numbers inequality $na^{n-1}b\leq(n-1)a^n+b^n$ by induction

Published on 23 Feb 2020 - 18:50

score 55 · Answer 2 · 2020-03-12 12:01:11

55

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How many Steiner Symmetrizations does it take to make an arbitrary set convex?

Published on 12 Mar 2020 - 12:01

score 176 · Answer 3 · 2020-03-22 15:00:22

176

Views

One of my old inequality (very sharp)

Published on 22 Mar 2020 - 15:00

score 119 · Answer 4 · 2020-03-23 03:14:16

119

Views

Inequality from Israel TST

Published on 23 Mar 2020 - 3:14

score 152 · Answer 5 · 2026-03-25 04:43:51

152

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For $a,b,c\in\left[\frac{1}{\sqrt{6}}, 6\right]$: $\sum_{cyc}\frac{4}{a+3b}\geq \sum_{cyc}\frac{3}{a+2b}$

Published on 25 Mar 2026 - 4:43

score 70 · Answer 6 · 2020-04-14 17:50:11

70

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Where is the flaw in this using Rearrangement Inequality?

Published on 14 Apr 2020 - 17:50

score 78 · Answer 7 · 2020-04-24 14:43:51

78

Views

$ \frac{a^{2}}{a+2 b^{3}}+\frac{b^{2}}{b+2 c^{3}}+\frac{c^{2}}{c+2 a^{3}} \geq 1 $

Published on 24 Apr 2020 - 14:43

score 139 · Answer 8 · 2020-05-01 05:22:06

139

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If $a+b+c=3$ Prove that $a^{2}+b^{2}+c^{2}\geq\frac{2+a}{2+b}+\frac{2+b}{2+c}+\frac{2+c}{2+a}$

Published on 01 May 2020 - 5:22

score 73 · Answer 9 · 2020-05-05 23:54:19

73

Views

Proving an inequality regarding increasing or decreasing sequences

Published on 05 May 2020 - 23:54

score 156 · Answer 10 · 2020-05-13 15:50:08

156

Views

If $a,b,c>0$ and $a+b+c=1$, then prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le 3+2\cdot\frac{a^3+b^3+c^3}{abc}$.

Published on 13 May 2020 - 15:50

score 714 · Answer 11 · 2020-06-03 18:26:36

714

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If $a, b, c\in\mathbb R^+, $ then prove that $a^3b+b^3c+c^3a\ge abc(a+b+c) .$

Published on 03 Jun 2020 - 18:26

score 58 · Answer 12 · 2020-06-12 13:15:50

58

Views

Maximum value of a parameter in a cyclic inequality?

Published on 12 Jun 2020 - 13:15

score 86 · Answer 13 · 2026-03-25 07:45:40

86

Views

If $abc=1$, then how do you prove $\frac{b-1}{bc+1}+\frac{c-1}{ac+1}+\frac{a-1}{ab+1} \geq 0$?

Published on 25 Mar 2026 - 7:45

score 146 · Answer 14 · 2020-06-22 02:06:52

146

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Prove: $\sqrt{\frac{bc}{a(3b+a)}} + \sqrt{\frac{ac}{b(3c+b)}} + \sqrt{\frac{ab}{c(3a+c)}} \ge \frac{3}{2}$.

Published on 22 Jun 2020 - 2:06

score 81 · Answer 15 · 2026-03-25 11:16:41

81

Views

on possible generalisations of $1\le\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\le 2$

Published on 25 Mar 2026 - 11:16

List Question

Natural numbers inequality $na^{n-1}b\leq(n-1)a^n+b^n$ by induction

How many Steiner Symmetrizations does it take to make an arbitrary set convex?

One of my old inequality (very sharp)

Inequality from Israel TST

For $a,b,c\in\left[\frac{1}{\sqrt{6}}, 6\right]$: $\sum_{cyc}\frac{4}{a+3b}\geq \sum_{cyc}\frac{3}{a+2b}$

Where is the flaw in this using Rearrangement Inequality?

$ \frac{a^{2}}{a+2 b^{3}}+\frac{b^{2}}{b+2 c^{3}}+\frac{c^{2}}{c+2 a^{3}} \geq 1 $

If $a+b+c=3$ Prove that $a^{2}+b^{2}+c^{2}\geq\frac{2+a}{2+b}+\frac{2+b}{2+c}+\frac{2+c}{2+a}$

Proving an inequality regarding increasing or decreasing sequences

If $a,b,c>0$ and $a+b+c=1$, then prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le 3+2\cdot\frac{a^3+b^3+c^3}{abc}$.

If $a, b, c\in\mathbb R^+, $ then prove that $a^3b+b^3c+c^3a\ge abc(a+b+c) .$

Maximum value of a parameter in a cyclic inequality?

If $abc=1$, then how do you prove $\frac{b-1}{bc+1}+\frac{c-1}{ac+1}+\frac{a-1}{ab+1} \geq 0$?

Prove: $\sqrt{\frac{bc}{a(3b+a)}} + \sqrt{\frac{ac}{b(3c+b)}} + \sqrt{\frac{ab}{c(3a+c)}} \ge \frac{3}{2}$.

on possible generalisations of $1\le\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\le 2$

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