Question

Find the smallest $\alpha$ such that, for all $x,y,z$, $\alpha\,\left(x^2-x+1\right)\left(y^2-y+1\right)\left(z^2-z+1\right)\ge(xyz)^2+|xyz|+1$.

score 137 · Answer 1 · 2016-07-12 08:17:50

137

Views

Find the smallest $\alpha$ such that, for all $x,y,z$, $\alpha\,\left(x^2-x+1\right)\left(y^2-y+1\right)\left(z^2-z+1\right)\ge(xyz)^2+|xyz|+1$.

Published on 12 Jul 2016 - 8:17

score 80 · Answer 2 · 2016-08-15 09:31:07

80

Views

inequality under condition x+y+z＝18xyz

Published on 15 Aug 2016 - 9:31

score 417 · Answer 3 · 2016-09-02 08:11:00

417

Views

Prove that $(1+a_1) \cdot (1+a_2) \cdot \dots \cdot (1+a_n) \geq 2^n$

Published on 02 Sep 2016 - 8:11

score 432 · Answer 4 · 2016-09-07 04:57:14

432

Views

How does this inequality follow from Hölder's inequality?

Published on 07 Sep 2016 - 4:57

score 907 · Answer 5 · 2016-10-21 13:31:25

907

Views

Generalization of Hölder's inequality with negative exponent

Published on 21 Oct 2016 - 13:31

score 112 · Answer 6 · 2016-11-17 04:50:38

112

Views

If $27 + abc = (a +b +c)(ab + ac+ bc)$ so $\frac{a^2}{a+2b} + \frac{b^2}{b+2c} + \frac{c^2}{c+2a }\geq \frac{3}{2}$

Published on 17 Nov 2016 - 4:50

score 141 · Answer 7 · 2016-11-19 13:36:40

141

Views

Showing $64\le\left(1+\frac1x\right)\left(1+\frac1y\right)\left(1+\frac1z\right)$

Published on 19 Nov 2016 - 13:36

score 328 · Answer 8 · 2017-01-10 08:09:34

328

Views

Inequality with square roots : $\frac{\sqrt{b+c}}{a}+\frac{\sqrt{c+a}}{b}+\frac{\sqrt{a+b}}{c}\geq \frac{4(a+b+c)}{\sqrt{(a+b)(b+c)(c+a)}}$

Published on 10 Jan 2017 - 8:09

score 739 · Answer 9 · 2026-03-25 09:29:06

739

Views

Proving $\left(\frac{a}{a+b+c}\right)^2+\left(\frac{b}{b+c+d}\right)^2+\left(\frac{c}{c+d+a}\right)^2+\left(\frac{d}{d+a+b}\right)^2\ge\frac{4}{9}$

Published on 25 Mar 2026 - 9:29

score 980 · Answer 10 · 2017-01-19 11:21:43

980

Views

How to Prove that $[(1+a)(1+b)(1+c)]^7$>$7^7a^4b^4c^4$

Published on 19 Jan 2017 - 11:21

score 1.6k · Answer 11 · 2017-01-25 18:54:09

1.6k

Views

Prove the inequality ${1 \over a^3 (b + c)} + {1 \over b^3 (a+c)} + {1 \over c^3 (a+b)} \ge \frac 32 $ given that $abc = 1$

Published on 25 Jan 2017 - 18:54

score 1.6k · Answer 12 · 2017-01-25 18:56:02

1.6k

Views

if : $abc=8 $ then : $(a+1)(b+1)(c+1)≥27$

Published on 25 Jan 2017 - 18:56

score 120 · Answer 13 · 2017-03-01 12:43:43

120

Views

Finding maximum of $x+y$

Published on 01 Mar 2017 - 12:43

score 100 · Answer 14 · 2017-03-10 22:48:57

100

Views

Prove the inequality, power.

Published on 10 Mar 2017 - 22:48

score 93 · Answer 15 · 2017-03-18 05:02:12

93

Views

Prove this inequality $\sqrt[3]{(a+b+c)(b+c+d)}\ge\sqrt[3]{ab}+\sqrt[3]{cd}$

Published on 18 Mar 2017 - 5:02

List Question

Find the smallest $\alpha$ such that, for all $x,y,z$, $\alpha\,\left(x^2-x+1\right)\left(y^2-y+1\right)\left(z^2-z+1\right)\ge(xyz)^2+|xyz|+1$.

inequality under condition x+y+z＝18xyz

Prove that $(1+a_1) \cdot (1+a_2) \cdot \dots \cdot (1+a_n) \geq 2^n$

How does this inequality follow from Hölder's inequality?

Generalization of Hölder's inequality with negative exponent

If $27 + abc = (a +b +c)(ab + ac+ bc)$ so $\frac{a^2}{a+2b} + \frac{b^2}{b+2c} + \frac{c^2}{c+2a }\geq \frac{3}{2}$

Showing $64\le\left(1+\frac1x\right)\left(1+\frac1y\right)\left(1+\frac1z\right)$

Inequality with square roots : $\frac{\sqrt{b+c}}{a}+\frac{\sqrt{c+a}}{b}+\frac{\sqrt{a+b}}{c}\geq \frac{4(a+b+c)}{\sqrt{(a+b)(b+c)(c+a)}}$

Proving $\left(\frac{a}{a+b+c}\right)^2+\left(\frac{b}{b+c+d}\right)^2+\left(\frac{c}{c+d+a}\right)^2+\left(\frac{d}{d+a+b}\right)^2\ge\frac{4}{9}$

How to Prove that $[(1+a)(1+b)(1+c)]^7$>$7^7a^4b^4c^4$

Prove the inequality ${1 \over a^3 (b + c)} + {1 \over b^3 (a+c)} + {1 \over c^3 (a+b)} \ge \frac 32 $ given that $abc = 1$

if : $abc=8 $ then : $(a+1)(b+1)(c+1)≥27$

Finding maximum of $x+y$

Prove the inequality, power.

Prove this inequality $\sqrt[3]{(a+b+c)(b+c+d)}\ge\sqrt[3]{ab}+\sqrt[3]{cd}$

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