$a^4+b^4+c^4+d^4=4\implies\sum\limits_{cyc}\frac{a^3}{bc}\geq4$ Let $a$, $b$, $c$ and $d$ be positive numbers such that $a^4+b^4+c^4+d^4=4$. Prove that:
$$\frac{a^3}{bc}+\frac{b^3}{cd}+\frac{c^3}{da}+\frac{d^3}{ab}\geq4$$
I tried C-S, BW and more, but without success. 
 A: We know that $$a^8 b^4 c^4 + b^8 c^4 d^4 + c^8 d^4 a^4+d^8 a^4 b^4 \leq 4.$$ (see this question for a proof)
By the generalized weighted mean inequality we have:
\begin{align}
\left(\frac{a^8 b^4 c^4 + b^8 c^4 d^4 + c^8 d^4 a^4+d^8 a^4 b^4}{a^4+b^4+c^4+d^4}\right)^{\frac{1}{4}} &\geq \frac{a^5bc+b^5cd+c^5da+d^5ab}{a^4+b^4+c^4+d^4}
\\\\ \Rightarrow 4 \left(\frac{a^8 b^4 c^4 + b^8 c^4 d^4 + c^8 d^4 a^4+d^8 a^4 b^4}{4}\right)^{\frac{1}{4}} &\geq a^5bc+b^5cd+c^5da+d^5ab
\\\\ \Rightarrow 4 &\geq a^5bc+b^5cd+c^5da+d^5ab
\\\\ \Rightarrow \frac{a^4+b^4+c^4+d^4}{a^5bc+b^5cd+c^5da+d^5ab} &\geq 1.
\end{align}
Let $f\colon \mathbb R_{> 0} \rightarrow \mathbb R$ be defined by $f(x):= \frac{1}{x}$. Since $f$ is convex, we have by Jensen's inequality:
\begin{align}
\frac{1}{4}\sum\limits_{cyc} \frac{a^3}{bc}
&=\sum\limits_{cyc} \frac{a^4}{a^4+b^4+c^4+d^4} f(a b c) 
\\&\geq f\left(\sum\limits_{cyc}\frac{a^5 b c}{a^4+b^4+c^4+d^4}\right) 
\\&= \frac{a^4+b^4+c^4+d^4}{a^5bc+b^5cd+c^5da+d^5ab}
\\&\geq 1.
\end{align}
A: I see a proof by  สนอง ห้วยเรไร:
We have:$$\sum_\limits{cyc}a\leq\sum_\limits{cyc}a^2$$and$$abc+bcd+cda+dab$$$$\leq ab+bc+cd+da$$$$\leq a^2+b^2+c^2+d^2\leq 4$$So:$$\sum_\limits{cyc}\frac{a^3}{bc}=\sum_\limits{cyc}\frac{a^4}{abc}$$$$\geq\frac{(\sum_\limits{cyc}a^2)^2}{\sum_\limits{cyc}abc}$$$$\geq\frac{(\sum_\limits{cyc}a^2)^2}{\sum_\limits{cyc}ab}$$$$\geq 4\frac{\sum_\limits{cyc}ab}{\sum_\limits{cyc}ab}=4$$
