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To prove:

$$(a^2+2)(b^2+2)(c^2+2)\ge9(ab+bc+ca)$$ where $a,b,c$ are positive real numbers. When does equality hold?

I thought equality would hold when $a = b = c$, but that doesn't seem to fit the statement.

To prove the inequality, I tried substituting $a = \tan A, b=\tan B$ and $c=\tan C$ in the hope of using some trigonometric manipulation. Also, my observational skills suggest that the LHS is the obvious cube of the GM of $3$ numbers; but that didn't help either.

The above inequality is same as proving $(a^2+b^2+c^2+6)^3≥81(ab+bc+ca)$

I know that $a^2+b^2+c^2\ge ab+bc+ca$, but I'm not able to take it from here.

Is this approach fine, or is there a shorter method for the problem? I doubt is $AM\ge GM$ is the most elegant way of arriving at the required statement. Please help.

Awnon Bhowmik
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stoic-santiago
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  • I feel like I saw this on AOPS before (I may have asked about it), and I'm suspicious of either Cauchy or Rearrangement. Where's @MichaelRozenberg when you need him? – Sean Roberson Dec 28 '17 at 04:12
  • Unlikely to be (directly) AM-GM since the condition of positiveness is not required. The equality case, however, does happen for certain $a=b=c$ values. – dxiv Dec 28 '17 at 04:35
  • Related : https://math.stackexchange.com/questions/834025/showing-prod-a22-geq-9-sum-ab – Arnaud D. Jun 07 '19 at 13:42

1 Answers1

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By C-S $$\left(\frac{(a+b)^2}{2}+1\right)(2+c^2)\geq(a+b+c)^2$$ and since $$(a^2+2)(b^2+2)\geq3\left(\frac{(a+b)^2}{2}+1\right)$$ it's $$(a-b)^2+2(ab-1)^2\geq0,$$ we obtain: $$(a^2+2)(b^2+2)(c^2+2)\geq3(a+b+c)^2\geq9(ab+ac+bc).$$

Michael Rozenberg
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