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Given $a,b,c>0$ and $(a+b)(b+c)(c+a)=8$. Show that $\displaystyle \frac {a+b+c} 3\geq\sqrt[27]{\frac{a^3+b^3+c^3}3}$.

Obviously, AM-GM seems to be suitable for LHS.

For RHS, $a^3+b^3+c^3=(a+b+c)^3-3(a+b)(b+c)(c+a)=(a+b+c)^3-24$, then I don't know what to do.

Can someone please teach me? Thank you.

p.s. That $\sqrt [27]{}$ is really terrible...

JSCB
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3 Answers3

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since $$(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(a+c)=a^3+b^3+c^3+24$$ so $$(a+b+c)^3=a^3+b^3+c^3+3+3+3+3+3+3+3+3\ge 9\sqrt[9]{(a^3+b^3+c^3)\times 3^8}$$

so $$\dfrac{a+b+c}{3}\ge\sqrt[27]{\dfrac{a^3+b^3+c^3}{3}}$$

math110
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  • Brilliant! ${}$ – JSCB Jul 04 '13 at 10:47
  • Ok suppose if we say take a=b=c=1 , then putting the values we get L.H.S always greater than $\\sqrt[27]{\frac{a^3+b^3+c^3}3}$ and for any values for a,b,c R.H.S will be less cause of $\\sqrt[27]{\frac{x}3}$ please don't mock..just a beginner! – joey rohan Jul 04 '13 at 12:15
  • @joeyrohan "... will be less cause of $\sqrt [27]{\frac{x}{3}}$" - What is $x$ here? – hola Nov 05 '14 at 18:40
  • can someone explain the 2nd last step? i mean where does that 9 and root 9 come from?? – Sherlock Homies Nov 06 '14 at 10:00
  • x s the value after substituting a,b,c in $$ \sqrt[]{(a^3+b^3+c^3)}$$ @pushpen.paul sorry for late response. – joey rohan Nov 09 '14 at 08:00
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Let $s = a+b+c$. Then, as you noted, the inequality is equivalent to: $$ \frac{s}{3} \geq \sqrt[27]{\frac{s^3-24}{3}} \qquad (\ast)$$ Because you know that $(a+b)(b+c)(c+a) = 8$, or equivalently $\frac{a+b}{2}\cdot\frac{b+c}{2}\cdot\frac{c+a}{2} = 1$, it follows that $s = \frac{a+b}{2} + \frac{b+c}{2} + \frac{c+a}{2} \geq 3$. In fact, this is all you can say about $s$, so our initial problem is equivalent to showing that $(\ast)$ holds for all $s \geq 3$. You can rewrite $(\ast)$ as: $$ \left( \frac{s}{3} \right)^2 - \frac{s^3-24}{3} \geq 0 \qquad (\ast\ast)$$ For $s = 3$ there is equality in $(\ast\ast)$, and we can in fact divide out $(s-3)^2$ from the left side. The resulting function: $$ \frac{\left( \frac{s}{3} \right)^2 - \frac{s^3-24}{3}}{(s-3)^2} $$ is a polynomial which takes strictly positive values for $s \geq 3$ (in fact, for $s \geq -3$). You can directly verify that by a variety of brute-force methods (possibly with help of some softwere like Mathematica)

A more satisfactory idea: Put $t = (s/3)^3 \geq 1$ for ease of notation. We can easily see that $(\ast\ast)$ is equivalent to $$t^{9} - 9t + 8 \geq 0 $$ But this follows directly by applying AM-GM to $t^9,1,1,1,1,1,1,1,1$!.

Jakub Konieczny
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I found a solution that uses a bit of calculus.

First using the identity you have already found, we figure the inequality is equivalent to $$\sqrt[3]{\frac{a^3+b^3+c^3+24}{27}} = \frac{a+b+c}{3} \le \sqrt[27]{\frac{a^3+b^3+c^3}{27}}.$$ Setting $x = \frac{a^3+b^3+c^3}{27}$ we find it equivalent to saying $$(x+\frac89)^9-x \ge 0.$$ This is where the calculus enters the equation. It is a standard method to prove that all the zeros of $p(x) = (x+\frac89)^9-x$ are in fact negative. Since $x\ge0$ by definition this proves the inequality. [The calculus: Take the derivative $p'(x) = 9(x+\frac89)^8-1$ which has two zeros $x_1,x_2$ smaller than zero. Show that $p(x_1),p(x_2)>0$.]

Peter Patzt
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