Question

Equal elementary symmetric polynomials implies existence of permutation

score 46 · Answer 1 · 2023-05-29 18:01:50

46

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Equal elementary symmetric polynomials implies existence of permutation

Published on 29 May 2023 - 18:01

score 120 · Answer 2 · 2023-06-12 15:06:41

120

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When does a finite (transitive) subgroup $G \leq S_n$ admit a "nice" resolvent invariant?

Published on 12 Jun 2023 - 15:06

score 183 · Answer 3 · 2026-03-26 04:35:16

183

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Prove: $\sqrt{\frac{a}{4a^2+bc}}+\sqrt{\frac{b}{4b^2+ac}}+\sqrt{\frac{c}{4c^2+ba}}\ge \frac{\sqrt{2}}{\sqrt{a+b+c}}.$

Published on 26 Mar 2026 - 4:35

score 51 · Answer 4 · 2026-03-26 02:55:45

51

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Basis of Symmetric Polynomials on Variables with Conditions

Published on 26 Mar 2026 - 2:55

score 213 · Answer 5 · 2023-06-27 07:07:59

213

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Find maximum and minimum of $\sum\limits_{k=0}^n a^kb^{n-k}$ for $a, b\ge 0$ with $a^2 + b^2 = 1$

Published on 27 Jun 2023 - 7:07

score 241 · Answer 6 · 2023-07-08 16:39:29

241

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Prove $2(a^2+b^2)(b^2+c^2)(c^2+a^2)\geq (a-b)^2(b-c)^2(c-a)^2$

Published on 08 Jul 2023 - 16:39

score 75 · Answer 7 · 2026-03-25 23:42:25

75

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Prove: $\sum\limits_{cyc}\frac{1}{b^2+bc+c^2} \ge \frac{2}{ab+bc+ca}+\frac{a+b+c}{3(a^3+b^3+c^3)}.$ with $a,b,c \ge 0 : ab+bc+ca>0.$

Published on 25 Mar 2026 - 23:42

score 64 · Answer 8 · 2023-07-09 14:38:00

64

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Generalizing inequalities on normalized elementary-symmetric polynomials

Published on 09 Jul 2023 - 14:38

score 38 · Answer 9 · 2026-03-28 09:44:14

38

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MacDonalds Hook content formula derivation

Published on 28 Mar 2026 - 9:44

score 81 · Answer 10 · 2026-03-26 06:10:54

81

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Finding $\small{\min\limits_{a+b+c=3}\sqrt{ab+bc+ca}\left(\frac1{\sqrt{bc+a}}+\frac1{\sqrt{ca+b}}+\frac1{\sqrt{ab+c}}\right).}$

Published on 26 Mar 2026 - 6:10

score 82 · Answer 11 · 2026-03-26 01:23:23

82

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a,b,c>0 and prove $\sum\frac{\sqrt{a+b}}{c+\sqrt{ab}}\ge \frac{3\sqrt{2}}{2}\sqrt{\frac{a+b+c}{ab+bc+ca}}.$

Published on 26 Mar 2026 - 1:23

score 108 · Answer 12 · 2026-03-25 12:50:03

108

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a,b,c>0 and prove $\frac{a}{a+\sqrt{(a+2b)(a+2c)}}+\frac{b}{b+\sqrt{(b+2a)(b+2c)}}+\frac{c}{c+\sqrt{(c+2b)(c+2a)}}\le \frac{3}{4}.$

Published on 25 Mar 2026 - 12:50

score 232 · Answer 13 · 2026-03-25 22:05:10

232

Views

Prove $\frac{1}{\sqrt{a+8b}}+\frac{1}{\sqrt{b+8c}}+\frac{1}{\sqrt{c+8a}}\ge 1$ for $ab + bc + ca = 3$

Published on 25 Mar 2026 - 22:05

score 62 · Answer 14 · 2026-03-25 12:52:55

62

Views

Proving Muirhead’s inequality by AM-GM

Published on 25 Mar 2026 - 12:52

score 82 · Answer 15 · 2026-03-25 22:05:11

82

Views

a+b+c=3. Prove that $\sqrt{(ab+2)(a+b)}+\sqrt{(bc+2)(b+c)}+\sqrt{(ca+2)(c+a)}\ge \frac{3\sqrt{3}}{2}\sqrt{3(ab+bc+ca)-abc}.$

Published on 25 Mar 2026 - 22:05

List Question

Equal elementary symmetric polynomials implies existence of permutation

When does a finite (transitive) subgroup $G \leq S_n$ admit a "nice" resolvent invariant?

Prove: $\sqrt{\frac{a}{4a^2+bc}}+\sqrt{\frac{b}{4b^2+ac}}+\sqrt{\frac{c}{4c^2+ba}}\ge \frac{\sqrt{2}}{\sqrt{a+b+c}}.$

Basis of Symmetric Polynomials on Variables with Conditions

Find maximum and minimum of $\sum\limits_{k=0}^n a^kb^{n-k}$ for $a, b\ge 0$ with $a^2 + b^2 = 1$

Prove $2(a^2+b^2)(b^2+c^2)(c^2+a^2)\geq (a-b)^2(b-c)^2(c-a)^2$

Prove: $\sum\limits_{cyc}\frac{1}{b^2+bc+c^2} \ge \frac{2}{ab+bc+ca}+\frac{a+b+c}{3(a^3+b^3+c^3)}.$ with $a,b,c \ge 0 : ab+bc+ca>0.$

Generalizing inequalities on normalized elementary-symmetric polynomials

MacDonalds Hook content formula derivation

Finding $\small{\min\limits_{a+b+c=3}\sqrt{ab+bc+ca}\left(\frac1{\sqrt{bc+a}}+\frac1{\sqrt{ca+b}}+\frac1{\sqrt{ab+c}}\right).}$

a,b,c>0 and prove $\sum\frac{\sqrt{a+b}}{c+\sqrt{ab}}\ge \frac{3\sqrt{2}}{2}\sqrt{\frac{a+b+c}{ab+bc+ca}}.$

a,b,c>0 and prove $\frac{a}{a+\sqrt{(a+2b)(a+2c)}}+\frac{b}{b+\sqrt{(b+2a)(b+2c)}}+\frac{c}{c+\sqrt{(c+2b)(c+2a)}}\le \frac{3}{4}.$

Prove $\frac{1}{\sqrt{a+8b}}+\frac{1}{\sqrt{b+8c}}+\frac{1}{\sqrt{c+8a}}\ge 1$ for $ab + bc + ca = 3$

Proving Muirhead’s inequality by AM-GM

a+b+c=3. Prove that $\sqrt{(ab+2)(a+b)}+\sqrt{(bc+2)(b+c)}+\sqrt{(ca+2)(c+a)}\ge \frac{3\sqrt{3}}{2}\sqrt{3(ab+bc+ca)-abc}.$

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