Question

Nice olympiad inequality :$\frac{xy^2}{4y^3+3}+\frac{yz^2}{4z^3+3}+\frac{zx^2}{4x^3+3}\leq \frac{3}{7}$

score 287 · Answer 1 · 2019-09-27 12:42:12

287

Views

Nice olympiad inequality :$\frac{xy^2}{4y^3+3}+\frac{yz^2}{4z^3+3}+\frac{zx^2}{4x^3+3}\leq \frac{3}{7}$

Published on 27 Sep 2019 - 12:42

score 107 · Answer 2 · 2026-03-23 10:29:25

107

Views

Show $ \frac {a^3}{a^2+b^2+c^2-bc}+\frac {b^3}{a^2+b^2+c^2-ca} +\frac {c^3}{a^2+b^2+c^2-ab} \le \frac {(a+b+c)^2 (a^4+b^4+c^4)}{6abc(ab+bc+ca)} $

Published on 23 Mar 2026 - 10:29

score 252 · Answer 3 · 2026-03-23 01:09:04

252

Views

How prove this inequality $\sum\limits_{cyc}\frac{a^3}{b+c+d}\ge \dfrac{1}{3}$,if $\sum\limits_{cyc}a\sqrt{bc}\ge 1$

Published on 23 Mar 2026 - 1:09

score 3.2k · Answer 4 · 2026-03-21 23:18:47

3.2k

Views

Proof of the inequality $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}$

Published on 21 Mar 2026 - 23:18

score 899 · Answer 5 · 2026-03-23 10:27:09

899

Views

Proving the inequality $4\ge a^2b+b^2c+c^2a+abc$

Published on 23 Mar 2026 - 10:27

score 3.6k · Answer 6 · 2026-03-23 07:11:19

3.6k

Views

If both $a,b>0$, then $a^ab^b \ge a^bb^a$

Published on 23 Mar 2026 - 7:11

score 294 · Answer 7 · 2019-10-11 14:58:34

294

Views

Inequality : $\frac{a}{\exp(a+b)}+\frac{b}{\exp(b+c)}+\frac{c}{\exp(c+a)}\leq \exp\Big(\frac{-2}{3}\Big)$

Published on 11 Oct 2019 - 14:58

score 314 · Answer 8 · 2019-10-12 16:21:05

314

Views

Typical Olympiad Inequality? If $\sum_i^na_i=n$ with $a_i>0$, then $\sum_{i=1}^n\left(\frac{a_i^3+1}{a_i^2+1}\right)^4\geq n$

Published on 12 Oct 2019 - 16:21

score 48 · Answer 9 · 2026-03-09 05:49:44

48

Views

An elegant but difficult inequality.

Published on 09 Mar 2026 - 5:49

score 135 · Answer 10 · 2019-11-03 15:41:00

135

Views

The inequality $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \frac{3}{2}$

Published on 03 Nov 2019 - 15:41

score 91 · Answer 11 · 2019-11-30 00:04:21

91

Views

Largest integer $k$ such that $\frac{a^{m+1}+b^{m+1}}{a^m+b^m}\geq\sqrt[k]{\frac{a^k+b^k}2}$

Published on 30 Nov 2019 - 0:04

score 39 · Answer 12 · 2026-03-18 16:52:29

39

Views

Rearrangement inequality with infinite variables

Published on 18 Mar 2026 - 16:52

score 111 · Answer 13 · 2020-01-13 02:11:25

111

Views

Proving an inequality involving fractions and square roots holds

Published on 13 Jan 2020 - 2:11

score 23 · Answer 14 · 2020-01-27 23:50:42

23

Views

Can I do this? Divide the terms inside $ (\frac {(m-a)^2}{2b} + d)^N $ by $d$?

Published on 27 Jan 2020 - 23:50

score 96 · Answer 15 · 2020-02-16 09:05:20

96

Views

$\frac{a}{a+2b+c}+\frac{b}{b+2c+a}+\frac{c}{c+2a+b}\geq\frac{3}{4}$

Published on 16 Feb 2020 - 9:05

List Question

Nice olympiad inequality :$\frac{xy^2}{4y^3+3}+\frac{yz^2}{4z^3+3}+\frac{zx^2}{4x^3+3}\leq \frac{3}{7}$

Show $ \frac {a^3}{a^2+b^2+c^2-bc}+\frac {b^3}{a^2+b^2+c^2-ca} +\frac {c^3}{a^2+b^2+c^2-ab} \le \frac {(a+b+c)^2 (a^4+b^4+c^4)}{6abc(ab+bc+ca)} $

How prove this inequality $\sum\limits_{cyc}\frac{a^3}{b+c+d}\ge \dfrac{1}{3}$,if $\sum\limits_{cyc}a\sqrt{bc}\ge 1$

Proof of the inequality $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}$

Proving the inequality $4\ge a^2b+b^2c+c^2a+abc$

If both $a,b>0$, then $a^ab^b \ge a^bb^a$

Inequality : $\frac{a}{\exp(a+b)}+\frac{b}{\exp(b+c)}+\frac{c}{\exp(c+a)}\leq \exp\Big(\frac{-2}{3}\Big)$

Typical Olympiad Inequality? If $\sum_i^na_i=n$ with $a_i>0$, then $\sum_{i=1}^n\left(\frac{a_i^3+1}{a_i^2+1}\right)^4\geq n$

An elegant but difficult inequality.

The inequality $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \frac{3}{2}$

Largest integer $k$ such that $\frac{a^{m+1}+b^{m+1}}{a^m+b^m}\geq\sqrt[k]{\frac{a^k+b^k}2}$

Rearrangement inequality with infinite variables

Proving an inequality involving fractions and square roots holds

Can I do this? Divide the terms inside $ (\frac {(m-a)^2}{2b} + d)^N $ by $d$?

$\frac{a}{a+2b+c}+\frac{b}{b+2c+a}+\frac{c}{c+2a+b}\geq\frac{3}{4}$

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