SOS proof for $\sum_{cyc}\frac{a^3}{bc}\ge a+b+c$ I need an SOS(sum of squares) proof for $$\sum_{cyc}\frac{a^3}{bc}\ge a+b+c$$
if $a,b,c>0$
I already have a am-gm proof but is there a way to use SOS.
Am-gm proof :
$\frac{a^3}{bc}+b+c\ge 3a$                        .....by(AM-GM ineq.)
thus $$\sum \frac{a^3}{bc}+2\sum a \ge 3\sum a$$
or $$\sum_{cyc}\frac{a^3}{bc}\ge a+b+c$$

Proof 1. We have
$$\sum \left(\frac{a^3}{bc}-a\right)  = \sum \left(\frac{a^2}{b} -2a + b \right) + \left (\frac{a^3}{4bc}+\frac{3b^3}{4ca}-\frac{b^2}{c}  \right )$$
$$= \sum \frac{(a-b)^2}{b}+\frac{(a^2+2ab+3b^2)(a-b)^2}{4abc} \geqslant 0.$$
Proof 2. We write inequality as
$$a^4+b^4+c^4-abc(a+b+c) \geqslant 0.$$
We have
$$a^4+b^4+c^4 - a^2b^2-b^2b^2-b^2c^2 = \sum \frac{(a-b)^2(a+b)^2}{2},$$
and
$$a^2b^2+b^2c^2+c^2a^2-abc(a+b+c) = \sum \frac{c^2(a-b)^2}{2}.$$
Thefore inequality equivalent to
$$\sum \frac{(a-b)^2[(a+b)^2+c^2]}{2} \geqslant 0.$$
