How do I show that $\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a \ge 4$ for $a^2 + b^2 + c^2 + d^2 = 4$? Let $a, b, c, d$ be positive real numbers such that $a^2 + b^2 + c^2 + d^2 = 4$, show that $$\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a \ge 4.$$
My try:
$$\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a \ge a + b + c + d,$$
yet $$a + b + c + d \le \sqrt{4(a^2 + b^2 + c^2 + d^2)} = 4.$$
Thus, direct application of Cauchy-Schwarz inequality is too weak. I tried other methods but with no significant progress:
$$(\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a)^2 \ge \frac {(a^{4/3} + b^{4/3} + c^{4/3})^3}{a^2 + b^2 + c^2 + d^2} = \frac {(a^{4/3} + b^{4/3} + c^{4/3})^3}4.$$
I also observed that
$$(\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a) + (\frac {a^2}c + \frac {b^2}d + \frac {c^2}a + \frac {d^2}b) + (\frac {a^2}d + \frac {b^2}a + \frac {c^2}b + \frac {d^2}c) + (\frac {a^2}a + \frac {b^2}b + \frac {c^2}c + \frac {d^2}d) = 4 (\frac 1a + \frac 1b + \frac 1c + \frac 1d) \ge 16,$$
since $$\frac 1a + \frac 1b + \frac 1c + \frac 1d \ge \sqrt{\frac {(1 + 1 + 1 + 1)^3}{a^2 + b^2 + c^2 + d^2}} = 4.$$
Now my work might seem stupid or off-topic here, but I provide it here because I wish any of these attempts will lead to a solution. Any hints will be appreciated.

(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.
