How do I show that $\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a \ge 4$ for $a^2 + b^2 + c^2 + d^2 = 4$?

(This is actually from the deleted answer to a different question, posted here with permission.)

From Cauchy-Schwarz: $$ \left(\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a \right) \left( a^2 b + b^2 c + c^2 d + c^2 a \right) \ge (a^2+b^2+c^2+d^2)^2 =16 $$ therefore it suffices to show that $$ \tag{*} \left( a^2 b + b^2 c + c^2 d + c^2 a \right) \le 4 $$ Using Cauchy-Schwarz again: $$ \left( a^2 b + b^2 c + c^2 d + c^2 a \right)^2 \le (a^ 2+ b^2 + c^2 + d^2)(a^2b^2 + b^2c^2 + c^ 2d^2 + d^2 a^2) \\ = 4 (a^2+c^2)(b^2+d^2) \\ \le 4 \left( \frac {a^2+b^2+c^2+d^2}{2} \right) ^2 = 16 $$ with AM-GM in the last step. From this $(*)$ follows.

Just another way is to use Holder : $$\left(\sum_{cyc} \frac{a^2}b \right)^2 \left( \sum_{cyc} a^2b^2\right) \geqslant \left( \sum_{cyc} a^2\right)^3=4^3$$

So it remains to show $$a^2b^2+b^2c^2+c^2d^2+d^2a^2 = (a^2+c^2)(b^2+d^2) \leqslant \frac14(a^2+b^2+c^2+d^2)^2 = 4 $$

This may also be shown using Jensen's inequality with $x \mapsto \frac{1}{\sqrt{x}}$ and weights $\frac{a^2}{4}$, $\frac{b^2}{4}$, $\frac{c^2}{4}$, $\frac{d^2}{4}$, which by the given assumption sum to one.

\begin{align*} \frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a &= 4 \left( \frac{a^2}{4}\frac{1}{\sqrt{b^2}} + \frac{b^2}{4}\frac{1}{\sqrt{c^2}} + \frac{c^2}{4}\frac{1}{\sqrt{d^2}}+ \frac{d^2}{4}\frac{1}{\sqrt{a^2}} \right) \\ &\geq 4 \left( \frac{1}{\sqrt{ \frac{a^2}{4} b^2 + \frac{b^2}{4} c^2 + \frac{c^2}{4} d^2 + \frac{d^2}{4} a^2} } \right) \end{align*}

So it suffices to show that \begin{equation} \frac{a^2}{4} b^2 + \frac{b^2}{4} c^2 + \frac{c^2}{4} d^2 + \frac{d^2}{4} a^2 \leq 1 \end{equation}

This is the same ending point as the other existing answers, which factor and use AM-GM to finish.

$$\frac{a^2}{4} b^2 + \frac{b^2}{4} c^2 + \frac{c^2}{4} d^2 + \frac{d^2}{4} a^2 = \frac{1}{4} (a^2+c^2)(b^2+d^2) \leq \frac{1}{4^2}(a^2+b^2+c^2+d^2)^2 = 1 $$