Question

Double integral of $(y-2)^2$ over $\{(x,y)\in\mathbb{R}^2\colon x^2 +y^2 \le 6x\}$

score 45 · Answer 1 · 2018-05-31 18:03:32

45

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Double integral of $(y-2)^2$ over $\{(x,y)\in\mathbb{R}^2\colon x^2 +y^2 \le 6x\}$

Published on 31 May 2018 - 18:03

score 150 · Answer 2 · 2026-03-22 10:01:27

150

Views

What tools one should use for inequalities?

Published on 22 Mar 2026 - 10:01

score 259 · Answer 3 · 2018-07-18 21:32:20

259

Views

$ \frac {3 x y z}{(x+y) (y+z) (z+x)} + \sum\limits_{cycl}^{} \left(\frac {x+y}{x+y+ 2 z}\right)^2 \ge \frac {9}{8}.$

Published on 18 Jul 2018 - 21:32

score 206 · Answer 4 · 2026-03-25 00:04:25

206

Views

Prove that $\sqrt{1-x^2} + \sqrt{1-y^2} + \sqrt{1-z^2}\geq 4\sqrt{\frac{3-(x^2+y^2+z^2)}{5+x^2+y^2+z^2}}$.

Published on 25 Mar 2026 - 0:04

score 355 · Answer 5 · 2026-03-21 20:44:03

355

Views

Inequality with $abc=2$

Published on 21 Mar 2026 - 20:44

score 1.6k · Answer 6 · 2026-03-25 00:02:54

1.6k

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Prove that: $xy\sqrt{z}+yz\sqrt{x}+zx\sqrt{y}\geq x+y+z$

Published on 25 Mar 2026 - 0:02

score 225 · Answer 7 · 2018-08-23 04:50:47

225

Views

If $x+y+z=3$ then $\sum x\sqrt{x^3+3y} \ge 6$

Published on 23 Aug 2018 - 4:50

score 789 · Answer 8 · 2026-03-22 13:57:52

789

Views

UVW method application of basic theorem

Published on 22 Mar 2026 - 13:57

score 92 · Answer 9 · 2018-10-05 10:43:59

92

Views

How to prove the given inequality using CS or AM GM HM inequality

Published on 05 Oct 2018 - 10:43

score 378 · Answer 10 · 2018-12-04 16:51:16

378

Views

Prove that $({a\over a+b})^3+({b\over b+c})^3+ ({c\over c+a})^3\geq {3\over 8}$

Published on 04 Dec 2018 - 16:51

score 327 · Answer 11 · 2019-01-04 15:44:15

327

Views

Prove $ \frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b} + \frac{36}{a + b + c} \geq 20 $

Published on 04 Jan 2019 - 15:44

score 143 · Answer 12 · 2019-01-14 15:34:51

143

Views

Prove $\Sigma_{cyc}(\frac{a}{b-c}-3)^4\ge193$

Published on 14 Jan 2019 - 15:34

score 362 · Answer 13 · 2019-02-03 07:19:08

362

Views

$\sqrt{24ab+25}+\sqrt{24bc+25}+\sqrt{24ca+25}\geq 21$ if $a+b+c=ab+bc+ca$?

Published on 03 Feb 2019 - 7:19

score 112 · Answer 14 · 2019-02-21 19:11:33

112

Views

Inequality with a+b+c=1 and $18(a^4+b^4+c^4)+6(a^2+b^2+c^2)+1\geq24(a^3+b^3+c^3)$

Published on 21 Feb 2019 - 19:11

score 135 · Answer 15 · 2026-03-22 17:41:45

135

Views

Inequalities, but working around an absolute value

Published on 22 Mar 2026 - 17:41

List Question

Double integral of $(y-2)^2$ over $\{(x,y)\in\mathbb{R}^2\colon x^2 +y^2 \le 6x\}$

What tools one should use for inequalities?

$ \frac {3 x y z}{(x+y) (y+z) (z+x)} + \sum\limits_{cycl}^{} \left(\frac {x+y}{x+y+ 2 z}\right)^2 \ge \frac {9}{8}.$

Prove that $\sqrt{1-x^2} + \sqrt{1-y^2} + \sqrt{1-z^2}\geq 4\sqrt{\frac{3-(x^2+y^2+z^2)}{5+x^2+y^2+z^2}}$.

Inequality with $abc=2$

Prove that: $xy\sqrt{z}+yz\sqrt{x}+zx\sqrt{y}\geq x+y+z$

If $x+y+z=3$ then $\sum x\sqrt{x^3+3y} \ge 6$

UVW method application of basic theorem

How to prove the given inequality using CS or AM GM HM inequality

Prove that $({a\over a+b})^3+({b\over b+c})^3+ ({c\over c+a})^3\geq {3\over 8}$

Prove $ \frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b} + \frac{36}{a + b + c} \geq 20 $

Prove $\Sigma_{cyc}(\frac{a}{b-c}-3)^4\ge193$

$\sqrt{24ab+25}+\sqrt{24bc+25}+\sqrt{24ca+25}\geq 21$ if $a+b+c=ab+bc+ca$?

Inequality with a+b+c=1 and $18(a^4+b^4+c^4)+6(a^2+b^2+c^2)+1\geq24(a^3+b^3+c^3)$

Inequalities, but working around an absolute value

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