Question

I can't find the maximum value of P because of some problems in my solution.

score 69 · Answer 1 · 2019-09-21 16:39:45

69

Views

I can't find the maximum value of P because of some problems in my solution.

Published on 21 Sep 2019 - 16:39

score 257 · Answer 2 · 2014-04-16 18:06:29

257

Views

If $a^2+b^2=1$ where $a,b>0$ then find the minimum value of $(a+b+{1\over{ab}})$

Published on 16 Apr 2014 - 18:06

score 625 · Answer 3 · 2014-05-18 08:53:08

625

Views

How to prove $\sqrt{\frac{ab}{2a^2+bc+ca}}+\sqrt{\frac{bc}{2b^2+ca+ab}}+\sqrt{\frac{ca}{2c^2+ab+bc}}\ge\frac{81}{2}\cdot\frac{abc}{(a+b+c)^3}$

Published on 18 May 2014 - 8:53

score 595 · Answer 4 · 2014-05-23 04:47:47

595

Views

Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\frac{3}{\sqrt{7}}$

Published on 23 May 2014 - 4:47

score 203 · Answer 5 · 2026-03-25 07:46:10

203

Views

Seeking concise proof: $\frac18(a^2+b^2)(b^2+c^2)(c^2+a^2)\ge\frac1{27}(ab+bc+ca)^3$, where $a$, $b$, $c$ are positive numbers

Published on 25 Mar 2026 - 7:46

score 464 · Answer 6 · 2014-08-10 14:13:42

464

Views

If the sum of cubes of $a,b,c,d$ is $1$, then $\frac{1}{1-bcd}+\frac{1}{1-cda}+\frac{1}{1-dab}+\frac{1}{1-abc}\le \frac{16}{3}$

Published on 10 Aug 2014 - 14:13

score 480 · Answer 7 · 2014-08-23 15:44:05

480

Views

Prove that $\sum\limits_{\mathrm{cyc}}{\frac{1}{(x+2y)^2}} \geq\frac{1}{xy+yz+zx}$ for $x, y, z > 0$

Published on 23 Aug 2014 - 15:44

score 1.4k · Answer 8 · 2014-10-05 19:36:32

1.4k

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Hard inequality $ (xy+yz+zx)\left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}\right)\ge\frac{9}{4} $

Published on 05 Oct 2014 - 19:36

score 921 · Answer 9 · 2026-03-20 04:29:19

921

Views

How prove this $x^3+y^3+z^3+3\ge 2(x^2+y^2+z^2)$

Published on 20 Mar 2026 - 4:29

score 291 · Answer 10 · 2014-11-19 06:24:09

291

Views

How prove this inequality$4(a^2+b^2+c^2)+9a^2b^2c^2\ge 21$

Published on 19 Nov 2014 - 6:24

score 187 · Answer 11 · 2014-12-24 06:22:58

187

Views

prove that $a^2 b^2 (a^2 + b^2 - 2) \ge (a + b)(ab - 1)$

Published on 24 Dec 2014 - 6:22

score 48 · Answer 12 · 2026-03-09 05:49:44

48

Views

An elegant but difficult inequality.

Published on 09 Mar 2026 - 5:49

score 120 · Answer 13 · 2019-11-05 10:18:02

120

Views

Finding maximum with two constraints

Published on 05 Nov 2019 - 10:18

score 168 · Answer 14 · 2019-11-06 01:41:12

168

Views

Prove $(\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x})^3+12\ge 13(x^3+y^3+z^3)$ for positives $xyz = 1$

Published on 06 Nov 2019 - 1:41

score 730 · Answer 15 · 2019-12-02 12:02:52

730

Views

If $x,y,z>0.$Prove: $(x+y+z) \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right) \geq9\sqrt[]\frac{x^2+y^2+z^2}{xy+yz+zx}$

Published on 02 Dec 2019 - 12:02

List Question

I can't find the maximum value of P because of some problems in my solution.

If $a^2+b^2=1$ where $a,b>0$ then find the minimum value of $(a+b+{1\over{ab}})$

How to prove $\sqrt{\frac{ab}{2a^2+bc+ca}}+\sqrt{\frac{bc}{2b^2+ca+ab}}+\sqrt{\frac{ca}{2c^2+ab+bc}}\ge\frac{81}{2}\cdot\frac{abc}{(a+b+c)^3}$

Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\frac{3}{\sqrt{7}}$

Seeking concise proof: $\frac18(a^2+b^2)(b^2+c^2)(c^2+a^2)\ge\frac1{27}(ab+bc+ca)^3$, where $a$, $b$, $c$ are positive numbers

If the sum of cubes of $a,b,c,d$ is $1$, then $\frac{1}{1-bcd}+\frac{1}{1-cda}+\frac{1}{1-dab}+\frac{1}{1-abc}\le \frac{16}{3}$

Prove that $\sum\limits_{\mathrm{cyc}}{\frac{1}{(x+2y)^2}} \geq\frac{1}{xy+yz+zx}$ for $x, y, z > 0$

Hard inequality $ (xy+yz+zx)\left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}\right)\ge\frac{9}{4} $

How prove this $x^3+y^3+z^3+3\ge 2(x^2+y^2+z^2)$

How prove this inequality$4(a^2+b^2+c^2)+9a^2b^2c^2\ge 21$

prove that $a^2 b^2 (a^2 + b^2 - 2) \ge (a + b)(ab - 1)$

An elegant but difficult inequality.

Finding maximum with two constraints

Prove $(\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x})^3+12\ge 13(x^3+y^3+z^3)$ for positives $xyz = 1$

If $x,y,z>0.$Prove: $(x+y+z) \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right) \geq9\sqrt[]\frac{x^2+y^2+z^2}{xy+yz+zx}$

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