Question

Prove that $\sum_\text{cyc}\frac{a}{b^2}\ge 3\sum_{cyc}\frac{1}{a^2}$ for $a,b,c>0$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$

score 126 · Answer 1 · 2019-04-30 06:26:49

126

Views

Prove that $\sum_\text{cyc}\frac{a}{b^2}\ge 3\sum_{cyc}\frac{1}{a^2}$ for $a,b,c>0$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$

Published on 30 Apr 2019 - 6:26

score 133 · Answer 2 · 2019-05-05 11:52:39

133

Views

Proving $ \sum_{cyc}^{} \frac {a(a^3+b^3)}{a^2+ab+b^2} \ge \frac{2}{3} (a^2+b^2+c^2)$ for $a, b, c > 0$

Published on 05 May 2019 - 11:52

score 102 · Answer 3 · 2019-05-05 18:39:38

102

Views

A hard inequality $(a^2-ab+b^2 )(b^2-bc+c^2 )(c^2-ca+a^2 ) + 11abc \leq 12$

Published on 05 May 2019 - 18:39

score 311 · Answer 4 · 2019-05-09 13:54:44

311

Views

Inequality for $a,b,c>0$ $\sum_{cyc}\sqrt{\frac{a^3}{14a^2+4b^2}}\leq \sum_{cyc}\sqrt{\frac{a+b}{36}}$

Published on 09 May 2019 - 13:54

score 199 · Answer 5 · 2019-05-11 03:52:43

199

Views

Given three positive numbers $a,b,c$ so that $8abc\geqq a+b+c+5$. Prove that $\frac{1}{a+2b}+\frac{1}{b+ 2c}+\frac{1}{c+2a}\leqq1$ .

Published on 11 May 2019 - 3:52

score 218 · Answer 6 · 2019-05-11 10:57:24

218

Views

Prove that $k=0$ is the only $constant$ so $(\sum\limits_{cyc}a^{3}-\sum\limits_{cyc}a^{2}b)-k(a-b)(a-c)(b+c)\geqq0$ .

Published on 11 May 2019 - 10:57

score 296 · Answer 7 · 2019-05-18 13:23:56

296

Views

Given three positve numbers $a,b,c$. Prove that $\sum\limits_{cyc}\frac{a}{\sqrt{b(a+b)}}\geqq \sum\limits_{cyc}\frac{a}{\sqrt{b(c+a)}}$ .

Published on 18 May 2019 - 13:23

score 149 · Answer 8 · 2019-05-26 10:51:00

149

Views

Prove $a^2b(a-b)+ b^2c(b-c)+ c^2a(c-a)\geq 0$

Published on 26 May 2019 - 10:51

score 137 · Answer 9 · 2019-05-28 01:08:13

137

Views

Given three postive numbers $a,b,c$ so that $a\geqq b\geqq c$. Prove that $\sum\limits_{cyc}\frac{a+bW}{aW+b}\geqq 3$ .

Published on 28 May 2019 - 1:08

score 85 · Answer 10 · 2019-06-01 19:40:27

85

Views

Prove $\sum_{cyc} a(a-b)(a-2b) \ge 0$.Where $a,b,c \ge 0$

Published on 01 Jun 2019 - 19:40

score 114 · Answer 11 · 2019-06-12 15:24:53

114

Views

show this inequality $(x+y)^3+(y+z)^3+(z+w)^3+(w+x)^3\ge 8(x^2y+y^2z+z^2w+w^2x)$

Published on 12 Jun 2019 - 15:24

score 315 · Answer 12 · 2019-06-26 05:26:59

315

Views

Given two positive numbers $b,\,c$. Prove $\left ( \frac{3}{b}- 1 \right )(3- b)^{2}+ \left ( \frac{b}{c}- 1 \right )(b- c)^{2}+ (c- 1)^{3}\geqq 0$ .

Published on 26 Jun 2019 - 5:26

score 182 · Answer 13 · 2019-07-02 08:53:43

182

Views

$\frac{3(a + b + c)^2}{25(ab + bc + ca)(a^2 + b^2 + c^2)} \le \sum_{cyc}\frac{1}{(3a + b + c)(c + 4a)} \le \frac{\sqrt{3(ab + bc + ca)}}{25abc}$

Published on 02 Jul 2019 - 8:53

score 127 · Answer 14 · 2019-07-24 12:53:49

127

Views

$\frac{x^2y}{z}+\frac{y^2z}{x}+\frac{xz^2}{y}\geq x^2+y^2+z^2$

Published on 24 Jul 2019 - 12:53

score 238 · Answer 15 · 2019-08-08 23:46:45

238

Views

Given $x+y+z=3, x,y,z>0 $ how to prove that $\frac{x}{y}+\frac{y}{z}+\frac{z}{x} >= x^2+y^2+z^2$

Published on 08 Aug 2019 - 23:46

List Question

Prove that $\sum_\text{cyc}\frac{a}{b^2}\ge 3\sum_{cyc}\frac{1}{a^2}$ for $a,b,c>0$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$

Proving $ \sum_{cyc}^{} \frac {a(a^3+b^3)}{a^2+ab+b^2} \ge \frac{2}{3} (a^2+b^2+c^2)$ for $a, b, c > 0$

A hard inequality $(a^2-ab+b^2 )(b^2-bc+c^2 )(c^2-ca+a^2 ) + 11abc \leq 12$

Inequality for $a,b,c>0$ $\sum_{cyc}\sqrt{\frac{a^3}{14a^2+4b^2}}\leq \sum_{cyc}\sqrt{\frac{a+b}{36}}$

Given three positive numbers $a,b,c$ so that $8abc\geqq a+b+c+5$. Prove that $\frac{1}{a+2b}+\frac{1}{b+ 2c}+\frac{1}{c+2a}\leqq1$ .

Prove that $k=0$ is the only $constant$ so $(\sum\limits_{cyc}a^{3}-\sum\limits_{cyc}a^{2}b)-k(a-b)(a-c)(b+c)\geqq0$ .

Given three positve numbers $a,b,c$. Prove that $\sum\limits_{cyc}\frac{a}{\sqrt{b(a+b)}}\geqq \sum\limits_{cyc}\frac{a}{\sqrt{b(c+a)}}$ .

Prove $a^2b(a-b)+ b^2c(b-c)+ c^2a(c-a)\geq 0$

Given three postive numbers $a,b,c$ so that $a\geqq b\geqq c$. Prove that $\sum\limits_{cyc}\frac{a+bW}{aW+b}\geqq 3$ .

Prove $\sum_{cyc} a(a-b)(a-2b) \ge 0$.Where $a,b,c \ge 0$

show this inequality $(x+y)^3+(y+z)^3+(z+w)^3+(w+x)^3\ge 8(x^2y+y^2z+z^2w+w^2x)$

Given two positive numbers $b,\,c$. Prove $\left ( \frac{3}{b}- 1 \right )(3- b)^{2}+ \left ( \frac{b}{c}- 1 \right )(b- c)^{2}+ (c- 1)^{3}\geqq 0$ .

$\frac{3(a + b + c)^2}{25(ab + bc + ca)(a^2 + b^2 + c^2)} \le \sum_{cyc}\frac{1}{(3a + b + c)(c + 4a)} \le \frac{\sqrt{3(ab + bc + ca)}}{25abc}$

$\frac{x^2y}{z}+\frac{y^2z}{x}+\frac{xz^2}{y}\geq x^2+y^2+z^2$

Given $x+y+z=3, x,y,z>0 $ how to prove that $\frac{x}{y}+\frac{y}{z}+\frac{z}{x} >= x^2+y^2+z^2$

Trending Questions

Popular # Hahtags

Popular Questions