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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 :

$$abc\left(a+\frac{bc}{a}\right)\left(b+\frac{ca}{b}\right)\left(c+\frac{ab}{c}\right)\geq abc(a+b)(b+c)(c+a) $$ 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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    $\begingroup$ The last inequality only holds when $c \geq a$. $\endgroup$ Dec 7, 2012 at 11:47
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    $\begingroup$ Also you may assume without loss of generality that $c \geq b \geq a$ to begin with. $\endgroup$
    – Cantor
    Dec 7, 2012 at 11:51

7 Answers 7

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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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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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By C-S $$\prod_{cyc}(a^2+bc)=\prod_{cyc}\sqrt{(a^2+bc)(ac+b^2)}\geq\prod_{cyc}\left(\sqrt{a^3c}+\sqrt{b^3c}\right)=abc\prod_{cyc}(a+b).$$ Done!

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Here is another proof using the Reverse Rearrangment Inequality.

Observe that with $ a \leq b \leq c$, we have $ bc \leq ca \leq ab $ and $a^2 \leq b^2 \leq c^2$. Hence, applying the RRI on these similarly ordered sequences, we conclude that

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

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  • $\begingroup$ A golden solution! Some orderings should be fixed, but it works! $\endgroup$
    – orangeskid
    Jan 31 at 19:34
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We have$:$ $$ \left( {a}^{2}+bc \right) \left( ac+{b}^{2} \right) \left( ab+{c}^{ 2} \right) -abc \left( a+b \right) \left( b+c \right) \left( c+a \right) $$

$$=ac \left( {b}^{2}+{c}^{2} \right) \left( a-b \right) ^{2}+ab \left( {a}^{2}+{c}^ {2} \right) \left( b-c \right) ^{2} +bc \left( { a}^{2}+{b}^{2} \right) \left( a-c \right) ^{2}\geqq 0$$ Done.

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  • $\begingroup$ Your proof looks nice to me; I suggest mention please the theorem involved in the proof. It looks like SOS. $\endgroup$ Oct 1, 2020 at 13:29
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$$LHS-RHS=abc\sum_{sym} a(a-b)(a-c)+\frac{1}{2}(ab+bc+ca)\left[c^2(a-b)^2+a^2(b-c)^2+b^2(c-a)^2\right]\ge 0$$

True by Schur 3 deg

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