Question

Prove that$\frac{1}{(x+1)^2+y^2+1}+\frac{1}{(y+1)^2+z^2+1}+\frac{1}{(z+1)^2+x^2+1}\le\frac{1}{2}$

score 85 · Answer 1 · 2019-11-17 11:09:10

85

Views

Prove that$\frac{1}{(x+1)^2+y^2+1}+\frac{1}{(y+1)^2+z^2+1}+\frac{1}{(z+1)^2+x^2+1}\le\frac{1}{2}$

Published on 17 Nov 2019 - 11:09

score 73 · Answer 2 · 2019-11-17 16:25:30

73

Views

Minimize $S=\frac{a}{b} + 2\sqrt{\frac{b}{c}+1} + 3\sqrt[3]{\frac{c}{a}+1}$ under $a=\max\{a,b,c\}$ and $a,b,c>0$.

Published on 17 Nov 2019 - 16:25

score 108 · Answer 3 · 2019-11-19 06:40:40

108

Views

The solution for $a+b+c+d = 4$ and $\left( \frac{1}{a^{12}} + ... + \frac{1}{d^{12}} \right) \left( 1 + 3abcd \right) = 16$

Published on 19 Nov 2019 - 6:40

score 203 · Answer 4 · 2019-11-19 11:36:11

203

Views

Proving the inequality $\sum_{\text{cyc}} \frac{a}{a^2+b^3+c^3}\le\frac1{5abc}$

Published on 19 Nov 2019 - 11:36

score 145 · Answer 5 · 2019-11-20 21:34:45

145

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Minimum value of $\frac{(x+8)^3}{x}$ without calculus, and weighted AM-GM inequality

Published on 20 Nov 2019 - 21:34

score 1.4k · Answer 6 · 2026-03-25 16:03:35

1.4k

Views

Prove that $(x + \sqrt[3]{abc})^3 \le (x + a)(x + b)(x + c) \le ( x + \frac{a + b + c}{3})^3$

Published on 25 Mar 2026 - 16:03

score 79 · Answer 7 · 2019-11-22 14:42:09

79

Views

Proving that $\sqrt{k^2 a^2+bc}+\sqrt{k^2 b ^2+ac}+\sqrt{k^2 c ^2+ab}\le(k+\frac12)\cdot(a+b+c)$

Published on 22 Nov 2019 - 14:42

score 51 · Answer 8 · 2019-11-22 18:35:37

51

Views

Let $x,y,z\geq 0$. Then $2(x+y)(x+z)(y+z)+2xyz\le(xy+yz+xz)^2+(x+y+z)^2$.

Published on 22 Nov 2019 - 18:35

score 141 · Answer 9 · 2026-03-27 02:53:16

141

Views

How do we prove this inequality?

Published on 27 Mar 2026 - 2:53

score 58 · Answer 10 · 2019-11-26 14:39:46

58

Views

Prove $4\sum_{cyc} \frac{a}{(a+b)^2}\le\frac1a+\frac1b+\frac1c$

Published on 26 Nov 2019 - 14:39

score 377 · Answer 11 · 2019-11-26 21:45:40

377

Views

How to solve a maximisation problem using the AM-GM inequality?

Published on 26 Nov 2019 - 21:45

score 71 · Answer 12 · 2026-03-17 01:32:21

71

Views

Prove that $\sum_\text{cyc} \frac{ab}{ab+b^2+ca}\le 1$

Published on 17 Mar 2026 - 1:32

score 69 · Answer 13 · 2026-03-27 01:00:12

69

Views

$ 27 + ( 2 + \frac{a^{2}}{bc} ) (2 + \frac{b^{2}}{ac}) (2 + \frac{c^{2}}{ab}) \ge 6 (a+b+c)(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) $

Published on 27 Mar 2026 - 1:00

score 47 · Answer 14 · 2019-11-29 13:10:51

47

Views

Inequality with some conditions involving Am-Gm

Published on 29 Nov 2019 - 13:10

score 77 · Answer 15 · 2019-12-04 18:34:57

77

Views

Show $\prod_{i=1}^n (1 - t + t a_i) \geq 1$ when $\prod_{i=1}^n a_i = 1$

Published on 04 Dec 2019 - 18:34

List Question

Prove that$\frac{1}{(x+1)^2+y^2+1}+\frac{1}{(y+1)^2+z^2+1}+\frac{1}{(z+1)^2+x^2+1}\le\frac{1}{2}$

Minimize $S=\frac{a}{b} + 2\sqrt{\frac{b}{c}+1} + 3\sqrt[3]{\frac{c}{a}+1}$ under $a=\max\{a,b,c\}$ and $a,b,c>0$.

The solution for $a+b+c+d = 4$ and $\left( \frac{1}{a^{12}} + ... + \frac{1}{d^{12}} \right) \left( 1 + 3abcd \right) = 16$

Proving the inequality $\sum_{\text{cyc}} \frac{a}{a^2+b^3+c^3}\le\frac1{5abc}$

Minimum value of $\frac{(x+8)^3}{x}$ without calculus, and weighted AM-GM inequality

Prove that $(x + \sqrt[3]{abc})^3 \le (x + a)(x + b)(x + c) \le ( x + \frac{a + b + c}{3})^3$

Proving that $\sqrt{k^2 a^2+bc}+\sqrt{k^2 b ^2+ac}+\sqrt{k^2 c ^2+ab}\le(k+\frac12)\cdot(a+b+c)$

Let $x,y,z\geq 0$. Then $2(x+y)(x+z)(y+z)+2xyz\le(xy+yz+xz)^2+(x+y+z)^2$.

How do we prove this inequality?

Prove $4\sum_{cyc} \frac{a}{(a+b)^2}\le\frac1a+\frac1b+\frac1c$

How to solve a maximisation problem using the AM-GM inequality?

Prove that $\sum_\text{cyc} \frac{ab}{ab+b^2+ca}\le 1$

$ 27 + ( 2 + \frac{a^{2}}{bc} ) (2 + \frac{b^{2}}{ac}) (2 + \frac{c^{2}}{ab}) \ge 6 (a+b+c)(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) $

Inequality with some conditions involving Am-Gm

Show $\prod_{i=1}^n (1 - t + t a_i) \geq 1$ when $\prod_{i=1}^n a_i = 1$

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