Question

Let $a+b+c=3,$ then prove$\sqrt[3]{\frac{a+b}{5ab+4}}+\sqrt[3]{\frac{c+b}{5cb+4}}+\sqrt[3]{\frac{a+c}{5ac+4}}\ge \sqrt[3]{6}$

score 466 · Answer 1 · 2023-08-08 10:03:10

466

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Let $a+b+c=3,$ then prove$\sqrt[3]{\frac{a+b}{5ab+4}}+\sqrt[3]{\frac{c+b}{5cb+4}}+\sqrt[3]{\frac{a+c}{5ac+4}}\ge \sqrt[3]{6}$

Published on 08 Aug 2023 - 10:03

score 69 · Answer 2 · 2023-08-08 12:55:32

69

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Finding $\small{\max\limits_{ab+bc+ca=abc+2}\frac{ab(2-c)}{a^2+abc+b^2}+\frac{bc(2-a)}{b^2+abc+c^2}+\frac{ca(2-b)}{c^2+abc+a^2}.}$

Published on 08 Aug 2023 - 12:55

score 115 · Answer 3 · 2026-02-23 08:05:04

115

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For $x+y+z=3,$ prove $\frac{1}{x^2+4}+\frac{1}{y^2+4}+\frac{1}{z^2+4}\le \frac{3}{5}.$

Published on 23 Feb 2026 - 8:05

score 108 · Answer 4 · 2026-03-25 18:50:37

108

Views

If $x,y,z>0: x+y+z=3,$ prove $\frac{x^2}{y^2+yz+z^2}+\frac{y^2}{x^2+xz+z^2}+\frac{z^2}{y^2+yx+x^2}\ge \frac{5}{3}(x^2+y^2+z^2)+2xyz-6.$

Published on 25 Mar 2026 - 18:50

score 82 · Answer 5 · 2023-08-10 01:48:27

82

Views

Find max $P=a^3b^2+b^3c^2+c^3a^2+7(ab+bc+ca).$

Published on 10 Aug 2023 - 1:48

score 40 · Answer 6 · 2023-08-10 06:34:44

40

Views

Find max $T={\sum\frac {1-2yz}{3\,{y}^{2}+3\,{z}^{2}+{x}^{2}}}.$

Published on 10 Aug 2023 - 6:34

score 130 · Answer 7 · 2023-08-10 10:02:08

130

Views

How to prove $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge \frac{3}{2}$ with pqr?

Published on 10 Aug 2023 - 10:02

score 66 · Answer 8 · 2023-08-11 06:41:27

66

Views

What is min $T=\sqrt{\frac{1-bc}{2b^2+5bc+2c^2}}+\sqrt{\frac{1-ac}{2a^2+5ac+2c^2}}+\sqrt{\frac{1-ba}{2b^2+5ba+2a^2}}$

Published on 11 Aug 2023 - 6:41

score 82 · Answer 9 · 2023-08-11 09:02:02

82

Views

Simple proof to find minimum $P=(a^3+1)(b^3+1)(c^3+1)$

Published on 11 Aug 2023 - 9:02

score 144 · Answer 10 · 2023-10-11 04:01:40

144

Views

$\frac34\le\sum_\text{cyc}\big(\frac{x}{x+1}\big)^2<2\ $ for $x,y,z>0$ and $xyz=1$.

Published on 11 Oct 2023 - 4:01

score 132 · Answer 11 · 2023-11-02 10:34:32

132

Views

Prove $\frac{a}{\sqrt{4a+3bc}}+\frac{b}{\sqrt{4b+3ca}}+\frac{c}{\sqrt{4c+3ab}}\le \frac{\sqrt{a+b+c+2}}{2}.$

Published on 02 Nov 2023 - 10:34

score 84 · Answer 12 · 2023-11-05 03:27:43

84

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Prove that $\frac{1}{\sqrt{4a+bc}}+\frac{1}{\sqrt{4b+ca}}+\frac{1}{\sqrt{4c+ab}}\le \sqrt{2(a+b+c)}.$

Published on 05 Nov 2023 - 3:27

score 111 · Answer 13 · 2023-11-10 12:11:58

111

Views

Prove $\sqrt{\frac{a+b}{c+ab}}+\sqrt{\frac{c+b}{a+cb}}+\sqrt{\frac{a+c}{b+ac}}\ge 3,$ if $ab+bc+ca=3.$

Published on 10 Nov 2023 - 12:11

score 74 · Answer 14 · 2023-11-11 16:02:14

74

Views

$\sum \frac{a^2+b^2}{c^2}-\frac{7}{2} \geq \sqrt[3]{\prod \left(\frac{a}{b+c}+\frac{1}{3}\right)}$ for a,b,c real

Published on 11 Nov 2023 - 16:02

score 136 · Answer 15 · 2023-11-15 23:35:23

136

Views

Given that $a,b,c>0$ and $abc=1$, prove that $a+b+c+\frac{3}{ab+bc+ca} \geq 4$

Published on 15 Nov 2023 - 23:35

List Question

Let $a+b+c=3,$ then prove$\sqrt[3]{\frac{a+b}{5ab+4}}+\sqrt[3]{\frac{c+b}{5cb+4}}+\sqrt[3]{\frac{a+c}{5ac+4}}\ge \sqrt[3]{6}$

Finding $\small{\max\limits_{ab+bc+ca=abc+2}\frac{ab(2-c)}{a^2+abc+b^2}+\frac{bc(2-a)}{b^2+abc+c^2}+\frac{ca(2-b)}{c^2+abc+a^2}.}$

For $x+y+z=3,$ prove $\frac{1}{x^2+4}+\frac{1}{y^2+4}+\frac{1}{z^2+4}\le \frac{3}{5}.$

If $x,y,z>0: x+y+z=3,$ prove $\frac{x^2}{y^2+yz+z^2}+\frac{y^2}{x^2+xz+z^2}+\frac{z^2}{y^2+yx+x^2}\ge \frac{5}{3}(x^2+y^2+z^2)+2xyz-6.$

Find max $P=a^3b^2+b^3c^2+c^3a^2+7(ab+bc+ca).$

Find max $T={\sum\frac {1-2yz}{3\,{y}^{2}+3\,{z}^{2}+{x}^{2}}}.$

How to prove $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge \frac{3}{2}$ with pqr?

What is min $T=\sqrt{\frac{1-bc}{2b^2+5bc+2c^2}}+\sqrt{\frac{1-ac}{2a^2+5ac+2c^2}}+\sqrt{\frac{1-ba}{2b^2+5ba+2a^2}}$

Simple proof to find minimum $P=(a^3+1)(b^3+1)(c^3+1)$

$\frac34\le\sum_\text{cyc}\big(\frac{x}{x+1}\big)^2<2\ $ for $x,y,z>0$ and $xyz=1$.

Prove $\frac{a}{\sqrt{4a+3bc}}+\frac{b}{\sqrt{4b+3ca}}+\frac{c}{\sqrt{4c+3ab}}\le \frac{\sqrt{a+b+c+2}}{2}.$

Prove that $\frac{1}{\sqrt{4a+bc}}+\frac{1}{\sqrt{4b+ca}}+\frac{1}{\sqrt{4c+ab}}\le \sqrt{2(a+b+c)}.$

Prove $\sqrt{\frac{a+b}{c+ab}}+\sqrt{\frac{c+b}{a+cb}}+\sqrt{\frac{a+c}{b+ac}}\ge 3,$ if $ab+bc+ca=3.$

$\sum \frac{a^2+b^2}{c^2}-\frac{7}{2} \geq \sqrt[3]{\prod \left(\frac{a}{b+c}+\frac{1}{3}\right)}$ for a,b,c real

Given that $a,b,c>0$ and $abc=1$, prove that $a+b+c+\frac{3}{ab+bc+ca} \geq 4$

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