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Let $a,b,c$ be three real positive(strictly) numbers. Prove that:

$$(a^2+bc)(b^2+ca)(c^2+ab) \geq abc(a+b)(b+c)(c+a).$$

I tried :

$$abc(a+\frac{bc}{a})(b+\frac{ca}{b})(c+\frac{ab}{c})\geq abc(a+b)(b+c)(c+a) $$ and now I want to try to prove that for example $$a+\frac{bc}{a} \geq a+b$$

but I don't know if is is a good idea.

Thanks:)

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1  
The last inequality only holds when $c \geq a$. –  Andrew Uzzell Dec 7 '12 at 11:47
2  
Also you may assume without loss of generality that $c \geq b \geq a$ to begin with. –  Cantor Dec 7 '12 at 11:51

3 Answers 3

up vote 3 down vote accepted

Here $\prod_{cyc}$ refers to the cyclic product of $x, y, z$. Let $a=x^2, b=y^2, c=z^2$ for positives $x, y, z$. Then by the Cauchy-Schwarz inequality, we have: $$LHS^2=\prod_{cyc}[(x^4+y^2z^2)(x^2z^2+y^4)]\ge \prod_{cyc}(x^3z+y^3z)^2=x^2y^2z^2\prod_{cyc}(x^3+y^3)^2$$ Then, by Power-Mean and AM-GM: $$x^2y^2z^2\prod_{cyc}(x^3+y^3)^2\ge x^2y^2z^2\prod_{cyc}[\frac{(x^2+y^2)^3}{2}]\ge x^2y^2z^2\prod_{cyc}[(x^2+y^2)^2xy]=RHS^2$$ Thus $LHS^2\ge RHS^2$, and both sides are positive so $LHS\ge RHS$ as desired.

Sidenote: There is also a proof by direct expansion in $a, b, c$: upon expanding and rearranging the inequality becomes $\sum_{cyc}(a-b)^2(\frac{a+b}{2})(c^3+abc)\ge 0$, which is clear.

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Here's another approach that is less intensive. Show that

$$ (a^2 + bc) ( b^2 + ca) \geq ab(c+b)(c+a) $$

This is equivalent to

$$ c (a-b)^2 (a+b) \geq 0 $$

Multiply the 3 cyclic versions of the first inequality, and you get your conclusion. Equality holds if and only if $a=b=c$.


Note: The factorization of the second line is easily guessed, by observing the equality holds when $ c = 0, a = b, a = - b$.

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Note that we have, from the Cauchy-Schwarz inequality, that $$(a^2+bc)(b+c)\geq \left(\sqrt{a^2b}+\sqrt{bc^2}\right)^2=b(c+a)^2.$$ Writing the analogous inequalities and multiplying leads to the desired result. Equality holds iff $a=b=c.\Box$

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