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Prove that $(a^3+b^3+c^3)(a^2+b^2+c^2)\ge(a+b+c)(a^2b^2+b^2c^2+c^2a^2)$, where a, b, c > 0.

I did LHS-RHS, broke parentheses and cancelled terms.

I think I might use $a^2+b^2+c^2\ge ab+bc+ca$. Then I only need to prove:

$a^5+b^5+c^5\gt abc(a^2+b^2+c^2)$.

renmom
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4 Answers4

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Hint:For your last inequality

Apply AM-GM like this $$3a^5+b^5+c^5\ge 5a^3bc$$

Arctic Char
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Albus Dumbledore
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  • Cool. I just need to do the same for b and c, then added them all up to obtain the last inequality. Thank you. – renmom Oct 30 '20 at 20:27
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Notice the last inequality is $a^5+b^5+c^5\geq abc(a^2+b^2+c^2)=a^3bc+ab^3c+abc^3$. One way to do it would be to quote Muirhead's Inequality directly, since $(5,0,0)\succ(3,1,1)$. Alternatively, using AM-GM, you obtain:

$$a^5+a^5+a^5+b^5+c^5\geq 5a^3bc$$ $$b^5+b^5+b^5+c^5+a^5\geq 5ab^3c$$ $$c^5+c^5+c^5+a^5+b^5\geq 5abc^3$$

I leave you to finish the proof using the above 3 inequalities if you do not want to use Muirhead's.

JC12
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  • Thank you. Understand, when adding them up, it's exactly what I needed. I'll learn about Muirhead's - great to know. – renmom Oct 30 '20 at 20:30
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As the inequality is homogeneous, assume: $a+b+c = 3$.
We know $$ (a^2 + b^2 + c^2)^2 \geqslant 3(a^2b^2 + b^2c^2 + c^2a^2) = (a+b+c)(a^2b^2 + b^2c^2 + c^2a^2) $$ We need to prove now: $$ a^3 + b^3 + c^3\geqslant a^2 + b^2 + c^2\quad \textrm{with}\quad a+b+c=3 $$ Which is $$ 3(a^3 + b^3 + c^3) \geqslant (a+b+c) (a^2 + b^2 + c^2) $$Obviously true.
For proof, after expansion, you can add $$ a^3 + b^3 + c^3 \geqslant a^2b + b^2c + c^2a\ \textrm{and}\ a^3 + b^3 + c^3 \geqslant ab^2 + bc^2 + ca^2 $$Proof of these are only by AM-GM.

Book Of Flames
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    Very glad to learn how to use the assumption $a+b+c=3$ in homogeneous inequality. Thanks for the clean writing. – renmom Oct 30 '20 at 21:02
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Yes, C-S helps here: $$\sum_{cyc}a^3\sum_{cyc}a^2=\frac{\sum\limits_{cyc}a^3\sum\limits_{cyc}a\sum\limits_{cyc}a^2}{\sum\limits_{cyc}a}\geq\frac{\left(\sum\limits_{cyc}a^2\right)^2\sum\limits_{cyc}a^2}{\sum\limits_{cyc}a}=$$ $$=\frac{\frac{1}{3}\left(\sum\limits_{cyc}a^2\right)^2\sum\limits_{cyc}1\sum\limits_{cyc}a^2}{\sum\limits_{cyc}a}\geq\frac{\sum\limits_{cyc}a^2b^2\left(\sum\limits_{cyc}a\right)^2}{\sum\limits_{cyc}a}=\sum\limits_{cyc}a^2b^2\sum\limits_{cyc}a.$$

Michael Rozenberg
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  • Thanks again, Michael. I tried to understand why you thought about adding $(a+b+c)$ in the first step. Is it to neutralize the power 3 down to power 2? Then from there, you split the 3 $a^2+b^2+c^2$ terms into 2+1, to match up the RHS. Magic. – renmom Oct 30 '20 at 21:23
  • @lilige Because I want to use C-S. I want to get $(a^3+b^3+c^3)(a+b+c)$ – Michael Rozenberg Oct 30 '20 at 21:37
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    Got it! Good trick for me to keep in mind. Cool. – renmom Oct 30 '20 at 21:39