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Given $a,b,c>0$ and $a+b+c=1$ then show that $$2+\frac{ab}{c} +\frac{bc}{a} +\frac{ca}{b} \leqslant \frac{1}{9abc}.$$

My work: the inequality is equivalent to $$18abc + 9((ab)^2 +(bc)^2 +(ca)^2 ) \leq 1.$$

Also, by AM-GM, gives $abc\leqslant \frac{1}{27}.$

Am I on the right track or how to proceed further?

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2 Answers 2

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Given $a,b,c>0$ and $a+b+c=1$.

Recall a well-known algebraic identity.

$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$

i.e.,

$$a^3+b^3+c^3-3abc=(a+b+c)^2-3(ab+bc+ca)=1-3(ab+bc+ca)$$

By the AM–GM inequality,

$$1-3(ab+bc+ca)\geq0\implies ab+bc+ca\leq\frac13$$

Now, $$\begin{aligned}2+\frac{ab}{c} +\frac{bc}{a} +\frac{ca}{b}&=\frac{2abc+a^2b^2+b^2c^2+c^2a^2}{abc}\\&=\frac{2abc+(ab+bc+ca)^2-2(ab\cdot bc+bc\cdot ca+ca\cdot ab)}{abc}\\&=\frac{2abc+(ab+bc+ca)^2-2abc(a+b+c)}{abc}\\&=\frac{(ab+bc+ca)^2}{abc}\\2+\frac{ab}{c} +\frac{bc}{a} +\frac{ca}{b}&\leq\frac{1}{9abc}\qquad\text{proved.} \end{aligned}$$

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Hint. After your first step, note that $$(ab+bc+ca)^2=2abc(a+b+c)+(ab)^2+(bc)^2+(ca)^2.$$ Therefore, recalling that $a+b+c=1$, it suffices to show that $9(ab+bc+ca)^2\leq 1$, that is $$ab+bc+ca\leq \frac{1}{3}.$$ Can you take it from here?

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