Question

Given four real numbers $a,b,c,d$ so that $1\leq a\leq b\leq c\leq d\leq 3$. Prove that $a^2+b^2+c^2+d^2\leq ab+ac+ad+bc+bd+cd.$

score 151 · Answer 1 · 2026-03-25 23:42:37

151

Views

Given four real numbers $a,b,c,d$ so that $1\leq a\leq b\leq c\leq d\leq 3$. Prove that $a^2+b^2+c^2+d^2\leq ab+ac+ad+bc+bd+cd.$

Published on 25 Mar 2026 - 23:42

score 144 · Answer 2 · 2026-03-27 14:03:23

144

Views

Prove that $1 \leq A \leq \frac{5}{4}$ and $0 \leq B < \frac{81}{16}$

Published on 27 Mar 2026 - 14:03

score 74 · Answer 3 · 2020-08-29 06:31:15

74

Views

Help with this proof of inequality

Published on 29 Aug 2020 - 6:31

score 144 · Answer 4 · 2026-03-25 19:04:47

144

Views

Prove that $\frac{1}{abc}+36\ge \frac{21}{ab+bc+ca}$

Published on 25 Mar 2026 - 19:04

score 160 · Answer 5 · 2026-03-26 16:05:14

160

Views

Symmetric Inequality for positive real numbers

Published on 26 Mar 2026 - 16:05

score 208 · Answer 6 · 2020-08-31 07:03:00

208

Views

Maximum value of $abc$ for $a+b+c=5$ and $a^2+b^2+c^2=11$

Published on 31 Aug 2020 - 7:03

score 138 · Answer 7 · 2020-08-31 08:19:36

138

Views

Finding the maximum value of $\sum_{cyc}\frac{1}{a^2-4a+9}$ s.t. $a+b+c =1$

Published on 31 Aug 2020 - 8:19

score 161 · Answer 8 · 2026-03-26 16:07:09

161

Views

Proving $\sum \frac{b+c}{9(a^2+3bc)+4(a+b+c)^2}\geqslant \frac{1}{4(a+b+c)}$

Published on 26 Mar 2026 - 16:07

score 124 · Answer 9 · 2026-03-26 16:10:14

124

Views

Proving $Q=\frac{a^{3}+b^{3}+c^{3}}{(a+b)(b+c)(c+a)}+k\cdot \frac{(ab+bc+ca)}{(a+b+c)^{2}}\geqslant \frac{3}{8}+\frac{k}{3}$

Published on 26 Mar 2026 - 16:10

score 160 · Answer 10 · 2026-03-26 21:26:24

160

Views

SOS proof for $\sum_{cyc}\frac{a^3}{bc}\ge a+b+c$

Published on 26 Mar 2026 - 21:26

score 225 · Answer 11 · 2020-09-02 11:19:53

225

Views

To prove $(a^2+b^2+c^2)(ab+bc+ca)^2 \ge (a^2+2bc)(b^2+2ca)(c^2+2ab)$.

Published on 02 Sep 2020 - 11:19

score 147 · Answer 12 · 2026-03-25 17:42:01

147

Views

Proving $4\Big(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \Big)+\frac{81}{(a+b+c)^2}\geqslant{\frac {7(a+b+c)}{abc}}$

Published on 25 Mar 2026 - 17:42

score 175 · Answer 13 · 2026-03-26 16:10:13

175

Views

Prove $\sum ab \sum \frac{1}{(a+b)^2} \geqslant \frac{9}{4}+\frac{kabc\sum (a^2-bc)}{(a+b+c)^3(ab+bc+ca)}$ for the best k.

Published on 26 Mar 2026 - 16:10

score 64 · Answer 14 · 2020-09-05 08:35:17

64

Views

Is there a way to convert a constrained double sum to unconstrained sum?

Published on 05 Sep 2020 - 8:35

score 206 · Answer 15 · 2026-03-26 23:09:36

206

Views

prove $a^3+b^3+c^3+3abc\ge \sum_{cyc}ab(a+b)$

Published on 26 Mar 2026 - 23:09

List Question

Given four real numbers $a,b,c,d$ so that $1\leq a\leq b\leq c\leq d\leq 3$. Prove that $a^2+b^2+c^2+d^2\leq ab+ac+ad+bc+bd+cd.$

Prove that $1 \leq A \leq \frac{5}{4}$ and $0 \leq B < \frac{81}{16}$

Help with this proof of inequality

Prove that $\frac{1}{abc}+36\ge \frac{21}{ab+bc+ca}$

Symmetric Inequality for positive real numbers

Maximum value of $abc$ for $a+b+c=5$ and $a^2+b^2+c^2=11$

Finding the maximum value of $\sum_{cyc}\frac{1}{a^2-4a+9}$ s.t. $a+b+c =1$

Proving $\sum \frac{b+c}{9(a^2+3bc)+4(a+b+c)^2}\geqslant \frac{1}{4(a+b+c)}$

Proving $Q=\frac{a^{3}+b^{3}+c^{3}}{(a+b)(b+c)(c+a)}+k\cdot \frac{(ab+bc+ca)}{(a+b+c)^{2}}\geqslant \frac{3}{8}+\frac{k}{3}$

SOS proof for $\sum_{cyc}\frac{a^3}{bc}\ge a+b+c$

To prove $(a^2+b^2+c^2)(ab+bc+ca)^2 \ge (a^2+2bc)(b^2+2ca)(c^2+2ab)$.

Proving $4\Big(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \Big)+\frac{81}{(a+b+c)^2}\geqslant{\frac {7(a+b+c)}{abc}}$

Prove $\sum ab \sum \frac{1}{(a+b)^2} \geqslant \frac{9}{4}+\frac{kabc\sum (a^2-bc)}{(a+b+c)^3(ab+bc+ca)}$ for the best k.

Is there a way to convert a constrained double sum to unconstrained sum?

prove $a^3+b^3+c^3+3abc\ge \sum_{cyc}ab(a+b)$

Trending Questions

Popular # Hahtags

Popular Questions