Question

For $a,b,c$ positive real numbers can it be true that: $(ab+bc+ca)^3 \ge (a^2+2b^2)(b^2+2c^2)(c^2+2a^2)$

score 82 · Answer 1 · 2016-05-08 01:44:33

82

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For $a,b,c$ positive real numbers can it be true that: $(ab+bc+ca)^3 \ge (a^2+2b^2)(b^2+2c^2)(c^2+2a^2)$

Published on 08 May 2016 - 1:44

score 82 · Answer 2 · 2016-05-09 00:39:51

82

Views

prove that $(ab+bc+ca)\bigg(\frac{a}{b(a^2+2b^2)}+\frac{b}{c(b^2+2c^2)}+\frac{c}{a(c^2+2a^2)}\bigg) \ge 3$

Published on 09 May 2016 - 0:39

score 182 · Answer 3 · 2016-05-12 16:59:00

182

Views

Cyclic Inequality in 3 variables

Published on 12 May 2016 - 16:59

score 160 · Answer 4 · 2016-05-14 09:37:10

160

Views

Let $a_1,...,a_{100}$ be non-negative numbers such that $a_1^2+...+a_{100}^2=1$ Prove that $a_1^2.a_2+...+a_{100}^2.a_1\le \frac{12}{25}$

Published on 14 May 2016 - 9:37

score 204 · Answer 5 · 2016-06-01 01:27:27

204

Views

Extreme values of $\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}$

Published on 01 Jun 2016 - 1:27

score 459 · Answer 6 · 2016-06-03 17:19:50

459

Views

Arithmetic Mean And Geometric Mean

Published on 03 Jun 2016 - 17:19

score 267 · Answer 7 · 2016-06-03 21:48:38

267

Views

How does $\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}$ relate to $\sqrt{x^2+y^2+z^2}$?

Published on 03 Jun 2016 - 21:48

score 672 · Answer 8 · 2016-06-09 23:10:50

672

Views

USA $2011$ contest inequality problem, proving $\frac{ab+1}{(a+b)^2}+\frac{bc+1}{(b+c)^2}+\frac{ca+1}{(c+a)^2}\ge 3$, under given condition.

Published on 09 Jun 2016 - 23:10

score 304 · Answer 9 · 2016-06-22 12:23:14

304

Views

Special Case of A.M-G.M Inequality.

Published on 22 Jun 2016 - 12:23

score 839 · Answer 10 · 2013-04-28 07:01:46

839

Views

Prove that $\frac{a_1^2}{a_1+a_2}+\frac{a_2^2}{a_2+a_3}+ \cdots \frac{a_n^2}{a_n+a_1} \geq \frac12$

Published on 28 Apr 2013 - 7:01

score 261 · Answer 11 · 2013-06-29 11:02:05

261

Views

If $abc=1$ and $\prod_{\text{cyc}}(a^2+b^2)=8$, then $(a+b+c)^3\prod_{\text{cyc}}(a^2+b^2)\ge\left (\sum_{\operatorname{cyc}} (a^2b+b^2a) \right) ^3$.

Published on 29 Jun 2013 - 11:02

score 114 · Answer 12 · 2026-03-25 12:29:22

114

Views

If $x,y,z>0$ are distinct and $x+y+z=1$ what is the minimum of $\left((1+x)(1+y)(1+z)\right)/\left((1-x)(1-y)(1-z)\right)$?

Published on 25 Mar 2026 - 12:29

score 250 · Answer 13 · 2026-03-25 18:52:29

250

Views

Evaluation Of Maximum Value Without Calculus

Published on 25 Mar 2026 - 18:52

score 669 · Answer 14 · 2013-09-25 01:13:50

669

Views

Prove that $\left(\sum^n_{k=1}x_k\right)\left(\sum^n_{k=1}y_k\right)\geq n^2$

Published on 25 Sep 2013 - 1:13

score 2.7k · Answer 15 · 2026-03-25 18:48:06

2.7k

Views

How prove this $a+b+c+3\sqrt[3]{abc}\ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})$

Published on 25 Mar 2026 - 18:48

List Question

For $a,b,c$ positive real numbers can it be true that: $(ab+bc+ca)^3 \ge (a^2+2b^2)(b^2+2c^2)(c^2+2a^2)$

prove that $(ab+bc+ca)\bigg(\frac{a}{b(a^2+2b^2)}+\frac{b}{c(b^2+2c^2)}+\frac{c}{a(c^2+2a^2)}\bigg) \ge 3$

Cyclic Inequality in 3 variables

Let $a_1,...,a_{100}$ be non-negative numbers such that $a_1^2+...+a_{100}^2=1$ Prove that $a_1^2.a_2+...+a_{100}^2.a_1\le \frac{12}{25}$

Extreme values of $\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}$

Arithmetic Mean And Geometric Mean

How does $\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}$ relate to $\sqrt{x^2+y^2+z^2}$?

USA $2011$ contest inequality problem, proving $\frac{ab+1}{(a+b)^2}+\frac{bc+1}{(b+c)^2}+\frac{ca+1}{(c+a)^2}\ge 3$, under given condition.

Special Case of A.M-G.M Inequality.

Prove that $\frac{a_1^2}{a_1+a_2}+\frac{a_2^2}{a_2+a_3}+ \cdots \frac{a_n^2}{a_n+a_1} \geq \frac12$

If $abc=1$ and $\prod_{\text{cyc}}(a^2+b^2)=8$, then $(a+b+c)^3\prod_{\text{cyc}}(a^2+b^2)\ge\left (\sum_{\operatorname{cyc}} (a^2b+b^2a) \right) ^3$.

If $x,y,z>0$ are distinct and $x+y+z=1$ what is the minimum of $\left((1+x)(1+y)(1+z)\right)/\left((1-x)(1-y)(1-z)\right)$?

Evaluation Of Maximum Value Without Calculus

Prove that $\left(\sum^n_{k=1}x_k\right)\left(\sum^n_{k=1}y_k\right)\geq n^2$

How prove this $a+b+c+3\sqrt[3]{abc}\ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})$

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