Question

Prove that $\frac{1}{\left(2a+b\right)^2}+\frac{1}{\left(2b+c\right)^2}+\frac{1}{\left(2c+a\right)^2}\ge\frac{1}{ab+bc+ca}$

score 239 · Answer 1 · 2017-03-02 15:01:47

239

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Prove that $\frac{1}{\left(2a+b\right)^2}+\frac{1}{\left(2b+c\right)^2}+\frac{1}{\left(2c+a\right)^2}\ge\frac{1}{ab+bc+ca}$

Published on 02 Mar 2017 - 15:01

score 220 · Answer 2 · 2017-03-08 13:17:00

220

Views

Square root inequality cyclic

Published on 08 Mar 2017 - 13:17

score 329 · Answer 3 · 2017-03-12 10:56:01

329

Views

If $abc=1$ so $\sum\limits_{cyc}\frac{7-6a}{2+a^2}\geq1$

Published on 12 Mar 2017 - 10:56

score 218 · Answer 4 · 2017-03-23 12:10:39

218

Views

Maximize $P=\frac{\sqrt{a^2-1}}{a}+\frac{\sqrt{b^2-1}}{b}+\frac{\sqrt{c^2-1}}{c} $

Published on 23 Mar 2017 - 12:10

score 248 · Answer 5 · 2017-03-29 14:34:30

248

Views

show that inequality $(x-3)^4+(y-3)^4+(z-3)^4\ge 193$

Published on 29 Mar 2017 - 14:34

score 125 · Answer 6 · 2023-03-11 14:24:18

125

Views

Maximize $\sum a_i^3$ for positive $a_i$'s with $\sum a_i=\sum a_i^2=n-1$

Published on 11 Mar 2023 - 14:24

score 3.3k · Answer 7 · 2023-05-16 13:10:53

3.3k

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How to prove: $ab+bc+ca+\sqrt[3]{abc}+20\ge 2\sum_{cyc}\sqrt{a(4a+ab+bc+ca)}$ if $a+b+c=5$

Published on 16 May 2023 - 13:10

score 750 · Answer 8 · 2023-05-20 11:03:02

750

Views

If $a,b,c\ge 0: ab+bc+ca=1$, find Min $P=\frac{1}{\sqrt{2a+bc}}+\frac{1}{\sqrt{2b+ca}}+\frac{1}{\sqrt{2c+ab}}$

Published on 20 May 2023 - 11:03

score 129 · Answer 9 · 2023-05-24 06:24:30

129

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$a,b,c\ge 0: ab+bc+ca=1$. Prove that: $\sqrt{a^2b^2+b^2c^2+c^2a^2}\ge \frac{a+b+c-6abc}{a^3+b^3+c^3+3abc}$

Published on 24 May 2023 - 6:24

score 466 · Answer 10 · 2023-06-26 07:00:32

466

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$a,b,c\ge 0: a+b+c+abc=4$. Prove that: $\sum_{cyc}\sqrt{\frac{a+b}{c+ab}}\ge 3.$

Published on 26 Jun 2023 - 7:00

score 74 · Answer 11 · 2023-07-09 08:27:39

74

Views

Prove: $\sum\limits_{cyc}\frac{1}{b^2+bc+c^2} \ge \frac{2}{ab+bc+ca}+\frac{a+b+c}{3(a^3+b^3+c^3)}.$ with $a,b,c \ge 0 : ab+bc+ca>0.$

Published on 09 Jul 2023 - 8:27

score 81 · Answer 12 · 2023-07-17 00:32:36

81

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a,b,c>0 and prove $\sum\frac{\sqrt{a+b}}{c+\sqrt{ab}}\ge \frac{3\sqrt{2}}{2}\sqrt{\frac{a+b+c}{ab+bc+ca}}.$

Published on 17 Jul 2023 - 0:32

score 231 · Answer 13 · 2023-07-20 02:55:37

231

Views

Prove $\frac{1}{\sqrt{a+8b}}+\frac{1}{\sqrt{b+8c}}+\frac{1}{\sqrt{c+8a}}\ge 1$ for $ab + bc + ca = 3$

Published on 20 Jul 2023 - 2:55

score 81 · Answer 14 · 2023-07-21 08:29:00

81

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a+b+c=3. Prove that $\sqrt{(ab+2)(a+b)}+\sqrt{(bc+2)(b+c)}+\sqrt{(ca+2)(c+a)}\ge \frac{3\sqrt{3}}{2}\sqrt{3(ab+bc+ca)-abc}.$

Published on 21 Jul 2023 - 8:29

score 140 · Answer 15 · 2023-07-24 02:59:44

140

Views

If $ab+bc+ca=3$, then prove that $\sum\sqrt{\frac{b+c}{bc+1}}\ge\frac{a+b+c+3}{\sqrt{a+b+c+abc}}$

Published on 24 Jul 2023 - 2:59

List Question

Prove that $\frac{1}{\left(2a+b\right)^2}+\frac{1}{\left(2b+c\right)^2}+\frac{1}{\left(2c+a\right)^2}\ge\frac{1}{ab+bc+ca}$

Square root inequality cyclic

If $abc=1$ so $\sum\limits_{cyc}\frac{7-6a}{2+a^2}\geq1$

Maximize $P=\frac{\sqrt{a^2-1}}{a}+\frac{\sqrt{b^2-1}}{b}+\frac{\sqrt{c^2-1}}{c} $

show that inequality $(x-3)^4+(y-3)^4+(z-3)^4\ge 193$

Maximize $\sum a_i^3$ for positive $a_i$'s with $\sum a_i=\sum a_i^2=n-1$

How to prove: $ab+bc+ca+\sqrt[3]{abc}+20\ge 2\sum_{cyc}\sqrt{a(4a+ab+bc+ca)}$ if $a+b+c=5$

If $a,b,c\ge 0: ab+bc+ca=1$, find Min $P=\frac{1}{\sqrt{2a+bc}}+\frac{1}{\sqrt{2b+ca}}+\frac{1}{\sqrt{2c+ab}}$

$a,b,c\ge 0: ab+bc+ca=1$. Prove that: $\sqrt{a^2b^2+b^2c^2+c^2a^2}\ge \frac{a+b+c-6abc}{a^3+b^3+c^3+3abc}$

$a,b,c\ge 0: a+b+c+abc=4$. Prove that: $\sum_{cyc}\sqrt{\frac{a+b}{c+ab}}\ge 3.$

Prove: $\sum\limits_{cyc}\frac{1}{b^2+bc+c^2} \ge \frac{2}{ab+bc+ca}+\frac{a+b+c}{3(a^3+b^3+c^3)}.$ with $a,b,c \ge 0 : ab+bc+ca>0.$

a,b,c>0 and prove $\sum\frac{\sqrt{a+b}}{c+\sqrt{ab}}\ge \frac{3\sqrt{2}}{2}\sqrt{\frac{a+b+c}{ab+bc+ca}}.$

Prove $\frac{1}{\sqrt{a+8b}}+\frac{1}{\sqrt{b+8c}}+\frac{1}{\sqrt{c+8a}}\ge 1$ for $ab + bc + ca = 3$

a+b+c=3. Prove that $\sqrt{(ab+2)(a+b)}+\sqrt{(bc+2)(b+c)}+\sqrt{(ca+2)(c+a)}\ge \frac{3\sqrt{3}}{2}\sqrt{3(ab+bc+ca)-abc}.$

If $ab+bc+ca=3$, then prove that $\sum\sqrt{\frac{b+c}{bc+1}}\ge\frac{a+b+c+3}{\sqrt{a+b+c+abc}}$

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