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Let a, b, c be positive numbers such that $abc=1$. Prove that $1+\frac{3}{a+b+c}\geq \frac{6}{ab+bc+ca}$

The usual methods do not seem to work, including a substitution $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$ and trying to apply Muirhead's inequality.

njguliyev
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Wilbur Smith
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  • Muirhead wouldn't work (directly) since the transformation makes the inequality non-symmetric. – Calvin Lin Sep 04 '13 at 20:59
  • I tried to prove the resulting non-symmetric inequality, without success. I also tried to not apply substitutions but get rid of the fractions and then homogenize the resulting inequality (since abc=1). Got nowhere. – Wilbur Smith Sep 04 '13 at 21:12

2 Answers2

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We have $(ab+bc+ca)^2 \geq 3 (a+b+c)abc $, which is equivalent to $ (ab-bc)^2 + (bc-ca)^2 + (ca-ab)^2 \geq 0$. Hence,

$$ 1 + \frac{3}{a+b+c} \geq 2 \sqrt{ \frac{3}{a+b+c} } \geq \frac{6}{ab+bc+ca}. $$

Calvin Lin
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$$\Longleftrightarrow (ab+bc+ac)+\dfrac{3(ab+bc+ac)}{a+b+c}\ge 6$$ since by $AM-GM$ inequality $$ (ab+bc+ac)+\dfrac{3(ab+bc+ac)}{a+b+c}\ge 2\sqrt{\dfrac{3(ab+bc+ac)^2}{a+b+c}}$$ $$\Longleftrightarrow2\sqrt{\dfrac{3(ab+bc+ac)^2}{a+b+c}}\ge 6$$ $$\Longleftrightarrow (ab+bc+ac)^2\ge 3(a+b+c)$$ since by $AM-GM$ $$(ab+bc+ac)^2\ge 3abc(a+b+c)=3(a+b+c)$$ By Done.

math110
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