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.
 A: (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.
A: 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 
$$a^2b^2+b^2c^2+c^2d^2+d^2a^2 = (a^2+c^2)(b^2+d^2)  \leqslant \frac14(a^2+b^2+c^2+d^2)^2 = 4 $$
A: 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.
$$\frac{a^2}{4} b^2 + \frac{b^2}{4} c^2 + \frac{c^2}{4} d^2 + \frac{d^2}{4} a^2 = \frac{1}{4} (a^2+c^2)(b^2+d^2)  \leq \frac{1}{4^2}(a^2+b^2+c^2+d^2)^2 = 1 $$
A: by the AM-GM inequality: 
Given that: 4= aˆ2 + bˆ2 + cˆ2 + dˆ2 ===> 1 = 1/4 (aˆ2 + bˆ2 + cˆ2 + dˆ2) ≥ 4^√((aˆ2)(bˆ2)(cˆ2)(dˆ2))= √(abcd)
since: 1≥ √(abcd)
(aˆ2+bˆ2+cˆ2+dˆ2)/4 (aˆ2/b + bˆ2/c + cˆ2/d + dˆ2/a)≥ 4ˆ√(abcd)(aˆ2 + bˆ2 + cˆ2 + dˆ2)ˆ4 = 4√(abcd)≥ 4 
Hence (aˆ2/b + bˆ2/c + cˆ2/d + dˆ2/a) ≥ 4
