Question

Prove $(a+b^2)(b+c^2)(c+a^2)\le13+abc(1-2abc)$ for $a + b + c = 3$

score 122 · Answer 1 · 2026-04-03 09:42:53

122

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Prove $(a+b^2)(b+c^2)(c+a^2)\le13+abc(1-2abc)$ for $a + b + c = 3$

Published on 03 Apr 2026 - 9:42

score 1.6k · Answer 2 · 2018-05-14 01:16:31

1.6k

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Find values of $p$ such that, for all $q>0$, $\frac{3(pq^2+p^2q+3q^2+3pq)}{p+q}>2p^2q$

Published on 14 May 2018 - 1:16

score 379 · Answer 3 · 2018-05-14 02:33:16

379

Views

Solving probability of inequalities with random variables

Published on 14 May 2018 - 2:33

score 371 · Answer 4 · 2018-05-14 07:35:57

371

Views

A tricky question involving inequalities

Published on 14 May 2018 - 7:35

score 54 · Answer 5 · 2018-05-14 15:39:23

54

Views

Proving Inequality in Integration

Published on 14 May 2018 - 15:39

score 207 · Answer 6 · 2018-05-14 16:26:44

207

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Does this inequality hold? $\sum \frac{a_i}{b_i} \leq \frac{\sum a_i}{\sum b_i}$

Published on 14 May 2018 - 16:26

score 52 · Answer 7 · 2026-03-25 01:26:59

52

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Help understand inequality algebra and stability analysis Du Fort-Frankel scheme PDE

Published on 25 Mar 2026 - 1:26

score 61 · Answer 8 · 2026-03-31 14:12:29

61

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How to solve the following simultaneous equations?

Published on 31 Mar 2026 - 14:12

score 53 · Answer 9 · 2018-05-15 03:41:51

53

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Is it true that $\{x\in \mathbb{R}: |1-f(x)|\leq 1-a\} = \{x\in \mathbb{R}: |f(x)|\geq a\}?$

Published on 15 May 2018 - 3:41

score 116 · Answer 10 · 2018-05-15 05:37:25

116

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If $\|f\|\leq 1,$ then is it true that $\{x\in\mathbb{R}: |1-f(x)|\leq 1-a\} = \{x\in\mathbb{R}: |f(x)|\geq a\}?$

Published on 15 May 2018 - 5:37

score 399 · Answer 11 · 2026-03-25 09:28:43

399

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Inequality $\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}>1$

Published on 25 Mar 2026 - 9:28

score 130 · Answer 12 · 2026-03-29 21:14:48

130

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Showing $(x^\top A^n x)(x^\top x)^{n-1}\geqslant(x^\top Ax)^n$ for any $n\in\mathbb N$ where $A$ is a symmetric matrix

Published on 29 Mar 2026 - 21:14

score 57 · Answer 13 · 2026-03-28 01:25:49

57

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Proof Verification: Using Taylor series to arrive at a certain inequality

Published on 28 Mar 2026 - 1:25

score 634 · Answer 14 · 2018-05-15 18:36:41

634

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Chebychev inequality generalization, essential supremum

Published on 15 May 2018 - 18:36

score 137 · Answer 15 · 2026-03-27 17:35:22

137

Views

Alternative way of thinking about irrationality of $\sqrt{2}$

Published on 27 Mar 2026 - 17:35

List Question

Prove $(a+b^2)(b+c^2)(c+a^2)\le13+abc(1-2abc)$ for $a + b + c = 3$

Find values of $p$ such that, for all $q>0$, $\frac{3(pq^2+p^2q+3q^2+3pq)}{p+q}>2p^2q$

Solving probability of inequalities with random variables

A tricky question involving inequalities

Proving Inequality in Integration

Does this inequality hold? $\sum \frac{a_i}{b_i} \leq \frac{\sum a_i}{\sum b_i}$

Help understand inequality algebra and stability analysis Du Fort-Frankel scheme PDE

How to solve the following simultaneous equations?

Is it true that $\{x\in \mathbb{R}: |1-f(x)|\leq 1-a\} = \{x\in \mathbb{R}: |f(x)|\geq a\}?$

If $\|f\|\leq 1,$ then is it true that $\{x\in\mathbb{R}: |1-f(x)|\leq 1-a\} = \{x\in\mathbb{R}: |f(x)|\geq a\}?$

Inequality $\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}>1$

Showing $(x^\top A^n x)(x^\top x)^{n-1}\geqslant(x^\top Ax)^n$ for any $n\in\mathbb N$ where $A$ is a symmetric matrix

Proof Verification: Using Taylor series to arrive at a certain inequality

Chebychev inequality generalization, essential supremum

Alternative way of thinking about irrationality of $\sqrt{2}$

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