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. 
 A: Just another way:
Let $x = a/b, y = b/c, z=c/a$, so that $xyz=1$.  Then we need to show $x^2+y^2+z^2\geqslant x+y+z$.  It is enough to show that $f(t)=t^2-t-\log t \geqslant 0$, which is easy as its only minimum is when $f(t=1)=0$.
A: Rewrite LHS as 6 terms, group them like this
$$
\left( \frac{1}{2} \cdot \frac{a^2}{b^2} + \frac{1}{2} \cdot \frac{b^2}{c^2} \right)+\dots
$$
Then apply AM-GM inequality
A: 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
$$(6,0,0)\succ(4,1,1).$$
A: After clearing the denominators we have to prove that
$$a^4c^2+b^4a^2+b^2c^4\geq abc(a^2c+ab^2+bc^2)$$
Now use that $$x^2+y^2+z^2\geq xy+yz+zx$$
This is $$(a^2c)^2+(bc^2)^2+(ab^2)^2\geq a^2cab^2+a^2cbc^2+ab^2bc^2=abc(a^2b+ac^2+b^2c)$$
Since$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=\frac{a^2c+ab^2+a^2b}{abc}$$
