Question

prove that $a^2(p-q)(p-r)+b^2(q-p)(q-r)+c^2(r-p)(r-q)\ge 0$

score 1k · Answer 1 · 2026-03-24 01:18:53

1k

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prove that $a^2(p-q)(p-r)+b^2(q-p)(q-r)+c^2(r-p)(r-q)\ge 0$

Published on 24 Mar 2026 - 1:18

score 179 · Answer 2 · 2026-03-24 06:26:47

179

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Proof of this simple inequality: $\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \geq 4$

Published on 24 Mar 2026 - 6:26

score 158 · Answer 3 · 2026-03-25 22:05:20

158

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Given $abc=1$ and $0< c \leq b \leq1\leq a$, prove that $8(a+b+c)^2\le9(1+a^2)(1+b^2)(1+c^2)$

Published on 25 Mar 2026 - 22:05

score 117 · Answer 4 · 2026-03-25 15:44:07

117

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Inequality $\frac{3}{\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}}\geq1+ \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$ for positive $a$, $b$, $c$

Published on 25 Mar 2026 - 15:44

score 886 · Answer 5 · 2026-03-24 00:04:49

886

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$ab(a+b) + bc(b+c) + ac(a+c) \geq \frac{2}{3}(a^{2}+b^{2}+c^{2})+ 4abc$ for $\frac1a+\frac1b+\frac1c=3$ and $a,b,c>0$

Published on 24 Mar 2026 - 0:04

score 159 · Answer 6 · 2026-03-24 10:38:00

159

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Inequality based on triangle: $\frac{3}{2}\le\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2$

Published on 24 Mar 2026 - 10:38

score 137 · Answer 7 · 2026-03-25 15:42:31

137

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Prove that $\frac{x^2y}{z}+\frac{y^2z}{x}+\frac{z^2x}{y} \geq x^2+y^2+z^2$

Published on 25 Mar 2026 - 15:42

score 130 · Answer 8 · 2026-03-24 15:57:09

130

Views

Prove that $(a+b+c)(ab+bc+ca)\geq9abc$ for $a, b, c \geq0.$

Published on 24 Mar 2026 - 15:57

score 3.2k · Answer 9 · 2026-03-24 06:57:11

3.2k

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Given $P(x)=x^{4}-4x^{3}+12x^{2}-24x+24,$ then $P(x)=|P(x)|$ for all real $x$

Published on 24 Mar 2026 - 6:57

score 125 · Answer 10 · 2026-03-23 23:57:36

125

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The least positive real number $k$ for which $k\left(\sqrt{(x+y)(y+z)(z+x)}\right)\geq x\sqrt{y}+y\sqrt{z}+z\sqrt{x}$

Published on 23 Mar 2026 - 23:57

score 145 · Answer 11 · 2026-03-24 11:41:42

145

Views

Show that $a^{bc}+b^{ca}+c^{ab}>2$ for $a,b,c\in(0,1)$

Published on 24 Mar 2026 - 11:41

score 180 · Answer 12 · 2026-03-23 17:23:56

180

Views

How prove this converse cauchy inequality of problem?

Published on 23 Mar 2026 - 17:23

score 151 · Answer 13 · 2026-03-25 21:51:08

151

Views

If $xy+xz+yz=3$ so $\sum\limits_{cyc}\left(x^2y+x^2z+2\sqrt{xyz(x^3+3x)}\right)\geq2xyz\sum\limits_{cyc}(x^2+2)$

Published on 25 Mar 2026 - 21:51

score 661 · Answer 14 · 2026-03-24 04:18:57

661

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Given $ab+bc+ca=3abc$, prove $\sqrt{\frac{a+b}{c(a^2+b^2 )}}+\sqrt{\frac{b+c}{a(b^2+c^2)}}+\sqrt{\frac{c+a}{b(c^2+a^2 )}}\leq 3$

Published on 24 Mar 2026 - 4:18

score 295 · Answer 15 · 2026-03-23 23:38:37

295

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If $a+b+c = 6$ and $a$, $b$, $c$ are nonnegative then $a^2+b^2+c^2 \geq 12$

Published on 23 Mar 2026 - 23:38

List Question

prove that $a^2(p-q)(p-r)+b^2(q-p)(q-r)+c^2(r-p)(r-q)\ge 0$

Proof of this simple inequality: $\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \geq 4$

Given $abc=1$ and $0< c \leq b \leq1\leq a$, prove that $8(a+b+c)^2\le9(1+a^2)(1+b^2)(1+c^2)$

Inequality $\frac{3}{\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}}\geq1+ \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$ for positive $a$, $b$, $c$

$ab(a+b) + bc(b+c) + ac(a+c) \geq \frac{2}{3}(a^{2}+b^{2}+c^{2})+ 4abc$ for $\frac1a+\frac1b+\frac1c=3$ and $a,b,c>0$

Inequality based on triangle: $\frac{3}{2}\le\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2$

Prove that $\frac{x^2y}{z}+\frac{y^2z}{x}+\frac{z^2x}{y} \geq x^2+y^2+z^2$

Prove that $(a+b+c)(ab+bc+ca)\geq9abc$ for $a, b, c \geq0.$

Given $P(x)=x^{4}-4x^{3}+12x^{2}-24x+24,$ then $P(x)=|P(x)|$ for all real $x$

The least positive real number $k$ for which $k\left(\sqrt{(x+y)(y+z)(z+x)}\right)\geq x\sqrt{y}+y\sqrt{z}+z\sqrt{x}$

Show that $a^{bc}+b^{ca}+c^{ab}>2$ for $a,b,c\in(0,1)$

How prove this converse cauchy inequality of problem?

If $xy+xz+yz=3$ so $\sum\limits_{cyc}\left(x^2y+x^2z+2\sqrt{xyz(x^3+3x)}\right)\geq2xyz\sum\limits_{cyc}(x^2+2)$

Given $ab+bc+ca=3abc$, prove $\sqrt{\frac{a+b}{c(a^2+b^2 )}}+\sqrt{\frac{b+c}{a(b^2+c^2)}}+\sqrt{\frac{c+a}{b(c^2+a^2 )}}\leq 3$

If $a+b+c = 6$ and $a$, $b$, $c$ are nonnegative then $a^2+b^2+c^2 \geq 12$

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