Question

show this inequality $\sum_{cyc}\frac{1}{5-2xy}\le 1$

score 117 · Answer 1 · 2020-05-04 15:09:45

117

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show this inequality $\sum_{cyc}\frac{1}{5-2xy}\le 1$

Published on 04 May 2020 - 15:09

score 136 · Answer 2 · 2020-05-04 22:32:17

136

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How can I approach this inequality?

Published on 04 May 2020 - 22:32

score 284 · Answer 3 · 2020-05-06 03:17:33

284

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Proving $(a^2+b^2+c^2)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}) +\frac{486(ab+bc+ca)^3}{(a+b+c)^6} \geqq 27$

Published on 06 May 2020 - 3:17

score 111 · Answer 4 · 2020-05-06 09:49:09

111

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Prove $\frac{3}{2} +\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \leqq \frac{a}{b}+\frac{b}{c} +\frac{c}{a}$

Published on 06 May 2020 - 9:49

score 147 · Answer 5 · 2026-03-25 09:43:13

147

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Prove $\frac{a^2}{(b+c)^2}+\frac{b^2}{(c+a)^2}+\frac{c^2}{(a+b)^2}+\frac{1}{4}\ge \frac{a^2+b^2+c^2}{ab+bc+ca}$

Published on 25 Mar 2026 - 9:43

score 182 · Answer 6 · 2026-03-25 09:39:54

182

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Alternative proofs of $a^2+b^2+c^2 \geqslant \frac{9abc}{a+b+c}+2(1+\sqrt 2)(a-b)(b-c)$

Published on 25 Mar 2026 - 9:39

score 156 · Answer 7 · 2020-05-13 15:50:08

156

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If $a,b,c>0$ and $a+b+c=1$, then prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le 3+2\cdot\frac{a^3+b^3+c^3}{abc}$.

Published on 13 May 2020 - 15:50

score 226 · Answer 8 · 2026-03-25 09:39:54

226

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Prove $\frac{a(b+c)}{b^2+bc+c^2} \geqslant \frac{8a^2+2a(b+c)-(b-c)^2}{3(a^2+b^2+c^2+ab+bc+ca)}.$

Published on 25 Mar 2026 - 9:39

score 141 · Answer 9 · 2026-02-23 04:22:29

141

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Prove $\sum \sqrt{{\frac {2{a}^{2}b}{a+c}}} \leqq a+b+c$ for $a,b,c>0$

Published on 23 Feb 2026 - 4:22

score 125 · Answer 10 · 2020-05-16 11:11:42

125

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Proving $(a+b+c)^2\prod_{cyc}(a+b)-4\sum_{cyc}(a^2b+a^2c)\sum_{cyc}ab\geqq 0$

Published on 16 May 2020 - 11:11

score 181 · Answer 11 · 2020-05-17 03:04:08

181

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Prove $\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b} +\frac{81abc}{4(a+b+c)^2} \geqq \frac{7}{4} (a+b+c)$

Published on 17 May 2020 - 3:04

score 131 · Answer 12 · 2026-03-25 08:13:37

131

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How to prove Fekete / Markov-Lukasz theorem: nonnegative univariate polynomial on [-1,1] can be decomposed accoording to even/odd-ness of degree

Published on 25 Mar 2026 - 8:13

score 137 · Answer 13 · 2020-05-20 12:10:52

137

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Prove $(a+b+c)^3 (a+b-c)(b+c-a)(c+a-b) \leqq 27a^2 b^ 2 c^2$

Published on 20 May 2020 - 12:10

score 198 · Answer 14 · 2020-05-21 00:44:21

198

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Prove $\frac{x^2+yz}{\sqrt{2x^2(y+z)}}+\frac{y^2+zx}{\sqrt{2y^2(z+x)}}+\frac{z^2+xy}{\sqrt{2z^2(x+y)}}\geqq 1$

Published on 21 May 2020 - 0:44

score 235 · Answer 15 · 2020-05-26 18:18:59

235

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Prove $\sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc + c^2} + \sqrt{c^2 + ac + a^2} \ge \sqrt{3}(a + b + c)$

Published on 26 May 2020 - 18:18

List Question

show this inequality $\sum_{cyc}\frac{1}{5-2xy}\le 1$

How can I approach this inequality?

Proving $(a^2+b^2+c^2)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}) +\frac{486(ab+bc+ca)^3}{(a+b+c)^6} \geqq 27$

Prove $\frac{3}{2} +\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \leqq \frac{a}{b}+\frac{b}{c} +\frac{c}{a}$

Prove $\frac{a^2}{(b+c)^2}+\frac{b^2}{(c+a)^2}+\frac{c^2}{(a+b)^2}+\frac{1}{4}\ge \frac{a^2+b^2+c^2}{ab+bc+ca}$

Alternative proofs of $a^2+b^2+c^2 \geqslant \frac{9abc}{a+b+c}+2(1+\sqrt 2)(a-b)(b-c)$

If $a,b,c>0$ and $a+b+c=1$, then prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le 3+2\cdot\frac{a^3+b^3+c^3}{abc}$.

Prove $\frac{a(b+c)}{b^2+bc+c^2} \geqslant \frac{8a^2+2a(b+c)-(b-c)^2}{3(a^2+b^2+c^2+ab+bc+ca)}.$

Prove $\sum \sqrt{{\frac {2{a}^{2}b}{a+c}}} \leqq a+b+c$ for $a,b,c>0$

Proving $(a+b+c)^2\prod_{cyc}(a+b)-4\sum_{cyc}(a^2b+a^2c)\sum_{cyc}ab\geqq 0$

Prove $\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b} +\frac{81abc}{4(a+b+c)^2} \geqq \frac{7}{4} (a+b+c)$

How to prove Fekete / Markov-Lukasz theorem: nonnegative univariate polynomial on [-1,1] can be decomposed accoording to even/odd-ness of degree

Prove $(a+b+c)^3 (a+b-c)(b+c-a)(c+a-b) \leqq 27a^2 b^ 2 c^2$

Prove $\frac{x^2+yz}{\sqrt{2x^2(y+z)}}+\frac{y^2+zx}{\sqrt{2y^2(z+x)}}+\frac{z^2+xy}{\sqrt{2z^2(x+y)}}\geqq 1$

Prove $\sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc + c^2} + \sqrt{c^2 + ac + a^2} \ge \sqrt{3}(a + b + c)$

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