Question

Prove the following inequality: $a^2+c^2-b^2>\frac{a^2c^2}{b^2}$

score 89 · Answer 1 · 2017-06-01 16:11:46

89

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Prove the following inequality: $a^2+c^2-b^2>\frac{a^2c^2}{b^2}$

Published on 01 Jun 2017 - 16:11

score 160 · Answer 2 · 2017-06-25 23:59:02

160

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If $a,b,c,d>0$ prove $\frac{2a^2+bc}{a+b+c} +\frac{2b^2+cd}{b+c+d}+\frac{2c^2+da}{c+d+a}+\frac{2d^2+ab}{d+a+b} \geq a+b+c+d$

Published on 25 Jun 2017 - 23:59

score 95 · Answer 3 · 2017-09-07 16:22:48

95

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Strange inequality over the positive reals

Published on 07 Sep 2017 - 16:22

score 455 · Answer 4 · 2017-09-15 08:12:30

455

Views

Turkevicius inequality

Published on 15 Sep 2017 - 8:12

score 130 · Answer 5 · 2017-10-28 10:38:53

130

Views

let $ \ \ 0<a \leq b \leq c \in \mathbb{R}$ then prove that :

Published on 28 Oct 2017 - 10:38

score 6.1k · Answer 6 · 2017-12-13 10:39:42

6.1k

Views

Prove the inequality $\frac{b+c}{a(y+z)}+\frac{c+a}{b(z+x)}+\frac{a+b}{c(x+y)}\geq \frac{3(a+b+c)}{ax+by+cz}$

Published on 13 Dec 2017 - 10:39

score 1.8k · Answer 7 · 2017-12-22 16:57:09

1.8k

Views

Elementary proof for $\sqrt[3]{xyz} \leq \dfrac{x+y+z}{3}$

Published on 22 Dec 2017 - 16:57

score 127 · Answer 8 · 2017-12-24 12:14:08

127

Views

Minimum value of rational expression

Published on 24 Dec 2017 - 12:14

score 118 · Answer 9 · 2018-01-01 20:16:18

118

Views

How to solve (in)equality of three variables with trigonometric solutions

Published on 01 Jan 2018 - 20:16

score 100 · Answer 10 · 2020-08-09 11:17:32

100

Views

Proving $\displaystyle \sum_{cyc}\frac{(a^2+b^2)}{a+b}\leqslant \frac{3(a^2+b^2+c^2)}{a+b+c}$

Published on 09 Aug 2020 - 11:17

score 294 · Answer 11 · 2020-08-14 23:24:54

294

Views

Proving $3(a^4 + b^4 + c^4 + d^4) + 4abcd \geq (a+b+c+d)(a^3 + b^3 + c^3 + d^3)$

Published on 14 Aug 2020 - 23:24

score 132 · Answer 12 · 2020-08-21 14:48:39

132

Views

Proving $\frac{a^3+b^3+c^3}{3}-abc\ge \frac{3}{4}\sqrt{(a-b)^2(b-c)^2(c-a)^2}$

Published on 21 Aug 2020 - 14:48

score 143 · Answer 13 · 2020-08-30 06:19:11

143

Views

Prove that $\frac{1}{abc}+36\ge \frac{21}{ab+bc+ca}$

Published on 30 Aug 2020 - 6:19

score 196 · Answer 14 · 2020-08-31 09:34:50

196

Views

Proving ${\frac {35{x}^{2}+7x(y+z)+23yz}{35(x^2+y^2+z^2)+37(xy+yz+zx)}}\leqslant \sqrt {{\frac {{x}^{2}+yz}{6\,{y}^{2} +6\,yz+6\,{z}^{2}}}}$

Published on 31 Aug 2020 - 9:34

score 146 · Answer 15 · 2020-09-03 02:25:15

146

Views

Proving $4\Big(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \Big)+\frac{81}{(a+b+c)^2}\geqslant{\frac {7(a+b+c)}{abc}}$

Published on 03 Sep 2020 - 2:25

List Question

Prove the following inequality: $a^2+c^2-b^2>\frac{a^2c^2}{b^2}$

If $a,b,c,d>0$ prove $\frac{2a^2+bc}{a+b+c} +\frac{2b^2+cd}{b+c+d}+\frac{2c^2+da}{c+d+a}+\frac{2d^2+ab}{d+a+b} \geq a+b+c+d$

Strange inequality over the positive reals

Turkevicius inequality

let $ \ \ 0<a \leq b \leq c \in \mathbb{R}$ then prove that :

Prove the inequality $\frac{b+c}{a(y+z)}+\frac{c+a}{b(z+x)}+\frac{a+b}{c(x+y)}\geq \frac{3(a+b+c)}{ax+by+cz}$

Elementary proof for $\sqrt[3]{xyz} \leq \dfrac{x+y+z}{3}$

Minimum value of rational expression

How to solve (in)equality of three variables with trigonometric solutions

Proving $\displaystyle \sum_{cyc}\frac{(a^2+b^2)}{a+b}\leqslant \frac{3(a^2+b^2+c^2)}{a+b+c}$

Proving $3(a^4 + b^4 + c^4 + d^4) + 4abcd \geq (a+b+c+d)(a^3 + b^3 + c^3 + d^3)$

Proving $\frac{a^3+b^3+c^3}{3}-abc\ge \frac{3}{4}\sqrt{(a-b)^2(b-c)^2(c-a)^2}$

Prove that $\frac{1}{abc}+36\ge \frac{21}{ab+bc+ca}$

Proving ${\frac {35{x}^{2}+7x(y+z)+23yz}{35(x^2+y^2+z^2)+37(xy+yz+zx)}}\leqslant \sqrt {{\frac {{x}^{2}+yz}{6\,{y}^{2} +6\,yz+6\,{z}^{2}}}}$

Proving $4\Big(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \Big)+\frac{81}{(a+b+c)^2}\geqslant{\frac {7(a+b+c)}{abc}}$

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