Question

If $ab+bc+ca=1,$ prove $1+36(abc)^2\ge\frac{21abc}{a+b+c}. $

score 109 · Answer 1 · 2023-08-03 05:08:54

109

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If $ab+bc+ca=1,$ prove $1+36(abc)^2\ge\frac{21abc}{a+b+c}. $

Published on 03 Aug 2023 - 5:08

score 70 · Answer 2 · 2023-08-03 13:11:26

70

Views

Prove $2(xy+yz+zx)+3\ge \sqrt{5x^2+4}+\sqrt{5y^2+4}+\sqrt{5z^2+4}$ when $x,y,z>0: x+y+z=3xyz.$

Published on 03 Aug 2023 - 13:11

score 188 · Answer 3 · 2023-08-03 14:09:26

188

Views

If $abc=25,$ find maximum $M=\frac{1}{b+c+12a}+\frac{1}{c+a+12b}+\frac{1}{a+b+12c}.$

Published on 03 Aug 2023 - 14:09

score 91 · Answer 4 · 2023-08-04 14:06:56

91

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Find the minimal constant $k$ :$\frac{a+k}{a+bc}+\frac{b+k}{b+ca}+\frac{c+k}{c+ab}\ge 3k+2,$ holds $ab+bc+ca=1.$

Published on 04 Aug 2023 - 14:06

score 79 · Answer 5 · 2023-08-05 03:44:56

79

Views

Prove $abc+\frac{1}{ab+bc+ca}\ge 1$ when $a+b+c=2.$

Published on 05 Aug 2023 - 3:44

score 119 · Answer 6 · 2023-08-05 10:21:54

119

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How to prove $\frac{5abc+1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1\right)+6\ge \sum_{\text{cyc}}\sqrt{5a^3+4}$ when $a^2+b^2+c^2=a+b+c$

Published on 05 Aug 2023 - 10:21

score 91 · Answer 7 · 2023-08-05 17:32:11

91

Views

Prove $\sum\dfrac{b+c}{a+bc}\ge 2\left[(a+b)(b+c)(c+a)+\frac{3abc}{a+b+c}\right]$ for $ab+bc+ca=1$

Published on 05 Aug 2023 - 17:32

score 143 · Answer 8 · 2023-08-06 03:16:48

143

Views

If $a+b+c=3,$ find max $T=\sqrt{\frac{6a+7bc}{6a+7}}+\sqrt{\frac{6b+7ca}{6b+7}}+\sqrt{\frac{6c+7ab}{6c+7}}$

Published on 06 Aug 2023 - 3:16

score 168 · Answer 9 · 2023-08-06 04:44:26

168

Views

$ab+bc+ca+abc=4$, prove $\sum\frac{2+\sqrt{ab}}{\sqrt{ab}+c}+\frac{a^2+b^2+c^2}{8abc}\ge\frac{39}{8}$

Published on 06 Aug 2023 - 4:44

score 41 · Answer 10 · 2023-08-06 09:13:11

41

Views

$(a^2+b^2+c^2+k(ab+bc+ca))(\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2})\geq \frac{9}{4}(2-k)$

Published on 06 Aug 2023 - 9:13

score 140 · Answer 11 · 2023-08-07 02:04:47

140

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If $x^2+y^2+z^2=1,$ prove $\frac{x(z^2-y^2+2y)}{z}+\frac{y(x^2-z^2+2z)}{x}+\frac{z(y^2-x^2+2x)}{y}\ge 2$

Published on 07 Aug 2023 - 2:04

score 187 · Answer 12 · 2026-03-25 15:42:13

187

Views

Finding $\small{\max\limits_{ab+bc+ca=3}\frac{ab}{2a+b}+\frac{bc}{2b+c}+\frac{ca}{2c+a}.}$

Published on 25 Mar 2026 - 15:42

score 118 · Answer 13 · 2023-08-07 13:24:55

118

Views

Finding $\small{\min\limits_{ab+bc+ca=3}\dfrac{1}{4a+bc}+\dfrac{1}{4b+ca}+\dfrac{1}{4c+ab}.}$

Published on 07 Aug 2023 - 13:24

score 188 · Answer 14 · 2026-03-23 03:32:23

188

Views

A difficult minimum

Published on 23 Mar 2026 - 3:32

score 66 · Answer 15 · 2023-08-08 03:04:14

66

Views

If $a^2+b^2+c^2=1$ prove $\frac{a^4-a^2}{bc-1}+\frac{b^4-b^2}{ca-1}+\frac{c^4-c^2}{ab-1}\le ab+bc+ca$

Published on 08 Aug 2023 - 3:04

List Question

If $ab+bc+ca=1,$ prove $1+36(abc)^2\ge\frac{21abc}{a+b+c}. $

Prove $2(xy+yz+zx)+3\ge \sqrt{5x^2+4}+\sqrt{5y^2+4}+\sqrt{5z^2+4}$ when $x,y,z>0: x+y+z=3xyz.$

If $abc=25,$ find maximum $M=\frac{1}{b+c+12a}+\frac{1}{c+a+12b}+\frac{1}{a+b+12c}.$

Find the minimal constant $k$ :$\frac{a+k}{a+bc}+\frac{b+k}{b+ca}+\frac{c+k}{c+ab}\ge 3k+2,$ holds $ab+bc+ca=1.$

Prove $abc+\frac{1}{ab+bc+ca}\ge 1$ when $a+b+c=2.$

How to prove $\frac{5abc+1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1\right)+6\ge \sum_{\text{cyc}}\sqrt{5a^3+4}$ when $a^2+b^2+c^2=a+b+c$

Prove $\sum\dfrac{b+c}{a+bc}\ge 2\left[(a+b)(b+c)(c+a)+\frac{3abc}{a+b+c}\right]$ for $ab+bc+ca=1$

If $a+b+c=3,$ find max $T=\sqrt{\frac{6a+7bc}{6a+7}}+\sqrt{\frac{6b+7ca}{6b+7}}+\sqrt{\frac{6c+7ab}{6c+7}}$

$ab+bc+ca+abc=4$, prove $\sum\frac{2+\sqrt{ab}}{\sqrt{ab}+c}+\frac{a^2+b^2+c^2}{8abc}\ge\frac{39}{8}$

$(a^2+b^2+c^2+k(ab+bc+ca))(\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2})\geq \frac{9}{4}(2-k)$

If $x^2+y^2+z^2=1,$ prove $\frac{x(z^2-y^2+2y)}{z}+\frac{y(x^2-z^2+2z)}{x}+\frac{z(y^2-x^2+2x)}{y}\ge 2$

Finding $\small{\max\limits_{ab+bc+ca=3}\frac{ab}{2a+b}+\frac{bc}{2b+c}+\frac{ca}{2c+a}.}$

Finding $\small{\min\limits_{ab+bc+ca=3}\dfrac{1}{4a+bc}+\dfrac{1}{4b+ca}+\dfrac{1}{4c+ab}.}$

A difficult minimum

If $a^2+b^2+c^2=1$ prove $\frac{a^4-a^2}{bc-1}+\frac{b^4-b^2}{ca-1}+\frac{c^4-c^2}{ab-1}\le ab+bc+ca$

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