<s>[INST] 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 $$\sum_{cyc}\frac{a^3}{bc}\ge a+b+c$$
 [/INST] Using SOS ... as requested.
\begin{eqnarray*}
(a^2-b^2)^2+(b^2-c^2)^2+(c^2-a^2)^2+2(a^2-bc)^2+2(b^2-ca)^2+2(c^2-ab)^2 \geq 0.
\end{eqnarray*}
Now divide by $4$ and we have
\begin{eqnarray*}
a^4+b^4+c^4 \geq abc(a+b+c).
\end{eqnarray*}
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