<s>[INST] How do we prove this inequality? Suppose $a,b,c > 0$. Prove that
$$\frac{a^2}{b^2} +\frac{b^2}{c^2} + \frac{c^2}{a^2} \geq \frac ab + \frac bc + \frac ca.$$
I've tried multiplying everything by the denominator and then I tried to use the rearrangement inequality, but it didn't yield the result I was looking for. I couldn't really think of anything else to do. 
 [/INST] Because by AM-GM we obtain: $$\sum_{cyc}\frac{a^2}{b^2}=\frac{1}{6}\sum_{cyc}\left(\frac{4a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)\geq\sum_{cyc}\sqrt[6]{\left(\frac{a^2}{b^2}\right)^4\cdot\frac{b^2}{c^2}\cdot\frac{c^2}{a^2}}=\sum_{cyc}\frac{a}{b}.$$
Just another way:
Let $x^3=\frac{a}{b}$, $y^3=\frac{b}{c}$ and $z^3=\frac{c}{a}$.
Thus, $xyz=1$ and we need to prove that
$$x^6+y^6+z^6\geq(x^3+y^3+z^3)xyz,$$ which is true by Muirhead because
$$(6,0,0)\succ(4,1,1).$$
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