AM-GM inequality: $\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{d} + \frac{d^2}{a} \geq a + b + c + d$ 
Let $a, b, c, d > 0$. Prove that $\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{d} + \frac{d^2}{a} \geq a + b + c + d$.

I'm supposed to prove this by AM-GM, but I can't see how. Any help would be appreciated.

Using this inequality is probably the quickest way
$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a}\geq \frac{(a+b+c+d)^2}{b+c+d+a}=a+b+c+d$$
Otherwise, using AM-GM we have:
$$\frac{a^2}{b}+b+\frac{b^2}{c}+c+\frac{c^2}{d}+d+\frac{d^2}{a}+a \geq 2\sqrt{\frac{a^2}{b}b}+2\sqrt{\frac{b^2}{c}c}+2\sqrt{\frac{c^2}{d}d}+2\sqrt{\frac{d^2}{a}a}=2(a+b+c+d)$$
