<s>[INST] 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
 [/INST] 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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