# show that if $a,b,c \in \mathbb{R}^+$

show that if $a,b,c \in \mathbb{R}^+$ different from zero, then:

$$(a^2+b^2+c^2)\cdot\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\leq(a+b+c)\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$$ I had no success in my attempts

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Just saying, you have $x,y,z$ and $a,b,c$ ;) – macydanim Dec 18 '12 at 20:10
@macydanim Thank you. – Henfe Dec 18 '12 at 20:12

## 2 Answers

$a=2012,b=c=1$.

Any other large number should work instead of $2012$.

If the inequality is reversed, just multiply and prove the following simple Lemma:

Lemma $f(x) =x+ \frac{1}{x}$ is increasing on $[1, \infty)$. In particular, for all $x\geq 1$ we have

$$f(x^2) \geq f(x) \,.$$

P.S. Maybe even simpler

$$\frac{a}{b}+\frac{b}{a} \geq 2 \Rightarrow \frac{a^2}{b^2}+\frac{b^2}{a^2}-\frac{a}{b}+\frac{b}{a}=(\frac{a}{b}+\frac{b}{a})^2-2-(\frac{a}{b}+\frac{b}{a}) \geq 2(\frac{a}{b}+\frac{b}{a} )-2-(\frac{a}{b}+\frac{b}{a}) \geq 0$$

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The direction of the inequality is flipped. Here is a proof of the correct direction. Upon expansion we see it suffices to prove: $$\frac{a^2}{b^2} + \frac{a^2}{c^2} + \frac{b^2}{a^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2} + \frac{c^2}{b^2} \ge \frac ab + \frac ac + \frac ba + \frac bc + \frac ca + \frac ca$$

Now note that for any positive real number $x$ we have $x^2 + \frac 1{x^2} \ge x + \frac 1x$ because the the function $f(x) = x + 1/x$ is monotonically increasing in $[1,\infty]$ and monotonically decreasing in $(0,1]$. Thus the result follows by applying this inequality on each of the terms of the above expression.

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