Question

Show that for every numbers a,b,c real positive we have $\sum a\frac{b^2+c^2}{b+c}\geq ab+bc+ca$

score 56 · Answer 1 · 2024-03-17 10:24:18

56

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Show that for every numbers a,b,c real positive we have $\sum a\frac{b^2+c^2}{b+c}\geq ab+bc+ca$

Published on 17 Mar 2024 - 10:24

score 48 · Answer 2 · 2026-03-26 12:37:17

48

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Equation $x^4+ax^3+bx^2+ax+1=0$ has at least 1 real root. $a,b \in \mathbb{R}$ What's the minimum value of $(a^2+b^2)$?

Published on 26 Mar 2026 - 12:37

score 46 · Answer 3 · 2024-03-30 01:18:15

46

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Finding Maximum Value using AM-GM Inequality

Published on 30 Mar 2024 - 1:18

score 103 · Answer 4 · 2024-03-31 11:57:52

103

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Given $x_1^3+x_2^3+...+x_9^3=0$. Find the maximum value of $S=x_1+x_2+...+ x_9$.

Published on 31 Mar 2024 - 11:57

score 91 · Answer 5 · 2015-03-11 18:30:47

91

Views

Weaker version of the $A.M.\;-\;G.M.$ inequality

Published on 11 Mar 2015 - 18:30

score 95 · Answer 6 · 2015-05-19 11:34:36

95

Views

Inequalities using AM-GM

Published on 19 May 2015 - 11:34

score 2.4k · Answer 7 · 2015-06-10 14:08:55

2.4k

Views

A.M.>G.M. of four numbers

Published on 10 Jun 2015 - 14:08

score 109 · Answer 8 · 2015-07-04 16:04:29

109

Views

Can AM–GM fail on a digital computer?

Published on 04 Jul 2015 - 16:04

score 749 · Answer 9 · 2026-03-26 14:22:18

749

Views

Prove the inequality $\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$ with the constraint $abc=1$

Published on 26 Mar 2026 - 14:22

score 495 · Answer 10 · 2015-08-26 05:44:16

495

Views

Russian MO 2004 Question involving the AM-GM inequality

Published on 26 Aug 2015 - 5:44

score 145 · Answer 11 · 2015-08-28 05:24:37

145

Views

Trouble understanding inequality proved using AM-GM inequality

Published on 28 Aug 2015 - 5:24

score 1.3k · Answer 12 · 2026-03-29 06:34:50

1.3k

Views

Cube of harmonic mean

Published on 29 Mar 2026 - 6:34

score 696 · Answer 13 · 2016-01-01 21:46:32

696

Views

Proving $ \frac{(a-b)^2}{8a}\leqslant\frac{a+b}{2} - \sqrt{ab} \leqslant\frac{(a-b)^2}{8b}$ when $ 0<b\le a$

Published on 01 Jan 2016 - 21:46

score 453 · Answer 14 · 2016-01-05 03:03:36

453

Views

Show that $\frac{a^2}{\sqrt{(1+a^3)(1+b^3)}}+\frac{b^2}{\sqrt{(1+b^3)(1+c^3)}}+\frac{c^2}{\sqrt{(1+c^3)(1+a^3)}} \geq \frac{4}{3}$

Published on 05 Jan 2016 - 3:03

score 248 · Answer 15 · 2016-01-07 17:11:55

248

Views

Prove that $\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}} \leq \frac{3}{2}$

Published on 07 Jan 2016 - 17:11

List Question

Show that for every numbers a,b,c real positive we have $\sum a\frac{b^2+c^2}{b+c}\geq ab+bc+ca$

Equation $x^4+ax^3+bx^2+ax+1=0$ has at least 1 real root. $a,b \in \mathbb{R}$ What's the minimum value of $(a^2+b^2)$?

Finding Maximum Value using AM-GM Inequality

Given $x_1^3+x_2^3+...+x_9^3=0$. Find the maximum value of $S=x_1+x_2+...+ x_9$.

Weaker version of the $A.M.\;-\;G.M.$ inequality

Inequalities using AM-GM

A.M.>G.M. of four numbers

Can AM–GM fail on a digital computer?

Prove the inequality $\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$ with the constraint $abc=1$

Russian MO 2004 Question involving the AM-GM inequality

Trouble understanding inequality proved using AM-GM inequality

Cube of harmonic mean

Proving $ \frac{(a-b)^2}{8a}\leqslant\frac{a+b}{2} - \sqrt{ab} \leqslant\frac{(a-b)^2}{8b}$ when $ 0<b\le a$

Show that $\frac{a^2}{\sqrt{(1+a^3)(1+b^3)}}+\frac{b^2}{\sqrt{(1+b^3)(1+c^3)}}+\frac{c^2}{\sqrt{(1+c^3)(1+a^3)}} \geq \frac{4}{3}$

Prove that $\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}} \leq \frac{3}{2}$

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