Nice Inequality

Question

I'm solving this inequality trying to use some changes of variable (for example $u=\frac{bc}{a}$, $v=\frac{ac}{b}$, $w=\frac{ab}{c}$), but I couldn't simplify the expression.

The inequality is: For $a,b,c>0$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$, show that $$\left(1+\frac{bc}{a}\right)\left(1+\frac{ac}{b}\right)\left(1+\frac{ab}{c}\right)\ge\frac{64}{81}(a+b+c)^2$$

Do you have any hint?

Thanks in advanced.

score 5 · Accepted Answer · answered May 12 '16 at 07:55

5

I don't know if this can help you:

$$\left(1+\frac{bc}{a}\right)\left(1+\frac{ac}{b}\right)\left(1+\frac{ab}{c}\right)\\ =\left(\frac{(a+bc)(b+ac)(c+ab)}{abc}\right)\\ =\left(\frac{(a+abc-ac-ab)(b+abc-bc-ab)(c+abc-bc-ac)}{abc}\right)\\ =(1+bc-c-b)(1+ac-c-a)(1+ab-b-a)\\ =(1-a)^2(1-b)^2(1-c)^2$$

answered May 12 '16 at 07:55

A s

708

Excellent!! With your observation is very easy to finish – sinbadh May 12 '16 at 08:11
+1. It reduces to prove $a+b+c \ge 9$, which is easy using $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ge \frac{9}{a+b+c}$. – f10w May 12 '16 at 09:01

Nice Inequality

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