Question

if $abc=1$, then $a^2+b^2+c^2\ge a+b+c$

score 9.4k · Answer 1 · 2014-04-05 08:22:16

9.4k

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if $abc=1$, then $a^2+b^2+c^2\ge a+b+c$

Published on 05 Apr 2014 - 8:22

score 283 · Answer 2 · 2014-04-25 11:06:11

283

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Proving AM-GM via $n \cdot (a_1^n + a_2^n + \dots + a_n^n) \ge (a_1^{n-1} + a_2^{n-1} + \dots + a_n^{n-1}) \cdot (a_1 + a_2 + \dots + a_n)$

Published on 25 Apr 2014 - 11:06

score 119 · Answer 3 · 2014-05-26 07:30:34

119

Views

Let $a,b,c>0$ such $ \frac{b+c}{a}+ \frac{c+a}{b}+ \frac{a+b}{c} = 2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac} \right)$

Published on 26 May 2014 - 7:30

score 135 · Answer 4 · 2014-07-14 10:42:08

135

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Proving Cauchy inequality involving four expression

Published on 14 Jul 2014 - 10:42

score 202 · Answer 5 · 2014-07-16 13:30:16

202

Views

Seeking concise proof: $\frac18(a^2+b^2)(b^2+c^2)(c^2+a^2)\ge\frac1{27}(ab+bc+ca)^3$, where $a$, $b$, $c$ are positive numbers

Published on 16 Jul 2014 - 13:30

score 172 · Answer 6 · 2019-10-23 21:42:42

172

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Generalization of the AM-GM inequality for three variables

Published on 23 Oct 2019 - 21:42

score 213 · Answer 7 · 2019-11-20 22:56:19

213

Views

Problematic inequality & hint

Published on 20 Nov 2019 - 22:56

score 139 · Answer 8 · 2026-03-08 05:48:21

139

Views

How do we prove this inequality?

Published on 08 Mar 2026 - 5:48

score 502 · Answer 9 · 2019-11-24 19:27:44

502

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Inequality in 4 variables Vasc's EV

Published on 24 Nov 2019 - 19:27

score 71 · Answer 10 · 2026-03-17 01:32:21

71

Views

Prove that $\sum_\text{cyc} \frac{ab}{ab+b^2+ca}\le 1$

Published on 17 Mar 2026 - 1:32

score 99 · Answer 11 · 2019-11-27 00:54:02

99

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Prove that $3 a^4+ 2 a^3 b - 3 a^2 b^2 - 2 a b^3 + 3 b^4\geq 0$.

Published on 27 Nov 2019 - 0:54

score 66 · Answer 12 · 2026-03-23 00:40:35

66

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Prove an inequality : $\sum_{cyc}\frac{a^3}{abu+b^2v}\geq \frac{a+b+c}{u+v}$ without Jensen's inequality

Published on 23 Mar 2026 - 0:40

score 300 · Answer 13 · 2026-03-23 03:52:03

300

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If $ab+bc+ca\ge1$, prove that $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{\sqrt{3}}{abc}$

Published on 23 Mar 2026 - 3:52

score 292 · Answer 14 · 2019-12-18 18:25:45

292

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

Published on 18 Dec 2019 - 18:25

score 205 · Answer 15 · 2019-12-21 20:46:33

205

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Proof verification for $x^{10}+y^{10}+z^{10}\ge x^9+y^9+z^9$ (where $xyz=1$ and $x,y,z\in \mathbb{R}^+$)

Published on 21 Dec 2019 - 20:46

List Question

if $abc=1$, then $a^2+b^2+c^2\ge a+b+c$

Proving AM-GM via $n \cdot (a_1^n + a_2^n + \dots + a_n^n) \ge (a_1^{n-1} + a_2^{n-1} + \dots + a_n^{n-1}) \cdot (a_1 + a_2 + \dots + a_n)$

Let $a,b,c>0$ such $ \frac{b+c}{a}+ \frac{c+a}{b}+ \frac{a+b}{c} = 2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac} \right)$

Proving Cauchy inequality involving four expression

Seeking concise proof: $\frac18(a^2+b^2)(b^2+c^2)(c^2+a^2)\ge\frac1{27}(ab+bc+ca)^3$, where $a$, $b$, $c$ are positive numbers

Generalization of the AM-GM inequality for three variables

Problematic inequality & hint

How do we prove this inequality?

Inequality in 4 variables Vasc's EV

Prove that $\sum_\text{cyc} \frac{ab}{ab+b^2+ca}\le 1$

Prove that $3 a^4+ 2 a^3 b - 3 a^2 b^2 - 2 a b^3 + 3 b^4\geq 0$.

Prove an inequality : $\sum_{cyc}\frac{a^3}{abu+b^2v}\geq \frac{a+b+c}{u+v}$ without Jensen's inequality

If $ab+bc+ca\ge1$, prove that $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{\sqrt{3}}{abc}$

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

Proof verification for $x^{10}+y^{10}+z^{10}\ge x^9+y^9+z^9$ (where $xyz=1$ and $x,y,z\in \mathbb{R}^+$)

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