<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] By Cauchy-Schwartz ineq.:
$$F=\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab} \ge \frac{(a^{3/2}+b^{3/2}+c^{3/2})^2}{ab+bc+ca}$$
Now, use Mean-Power ineq.: $M_{3/2} \ge M_1:$
$$F \ge \frac{3}{ab+bc+ca} \left(\frac{(a+b+c)^{3/2}}{3} \right)^2=\frac{(a+b+c)^3}{3(+bc+ca)}$$
Finally use $(a+b+c)^2 \ge 3(ab+bc+ca).$
Hence, $$F\ge (a+b+c)$$
</s>