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 $$\sum_{cyc}\frac{a^3}{bc}\ge a+b+c$$
Best Answer
The idea of the SOS's proof it's the following.
Let $P$ be a symmetric function of three variables $a$, $b$ and $c$ and let we can get: $$P(a,b,c)=\sum_{cyc}((a-b)Q(a,b,c)-(c-a)Q(a,c,b)).$$ Thus, $$P(a,b,c)=\sum_{cyc}((a-b)Q(a,b,c)-(c-a)Q(a,c,b))=$$ $$=\sum_{cyc}((a-b)Q(a,b,c)-(a-b)Q(b,a,c))=\sum_{cyc}(a-b)(Q(a,b,c)-Q(b,a,c))$$ and if $Q$ is a rational function we obtain a factor $a-b$ again.
There are some expressions, which we need to learn:
$$2a-b-c=a-b-(c-a),$$ $$a^2-bc=\frac{1}{2}((a-b)(a+c)-(c-a)(a+b))$$ and more similar.
This idea helps to prove inequalities by SOS without computer.
I hope now it's clear, how it works: $$\sum_{cyc}\frac{a^3}{bc}-\sum_{cyc}a=\sum_{cyc}\frac{a^3-abc}{bc}=\frac{1}{2}\sum_{cyc}\tfrac{a((a-b)(a+c)-(c-a)(a+b))}{bc}=$$ $$=\frac{1}{2}\sum_{cyc}(a-b)\left(\frac{a(a+c)}{bc}-\frac{b(b+c)}{ca}\right)=\frac{1}{2}\sum_{cyc}\tfrac{(a-b)^2(a^2+b^2+ab+ac+bc)}{abc}\geq0.$$ We saw before that we can get the expession $a^2-bc$ and after this we ended the proof.
Another example.
Let we need to prove the Nessbitt: $$\sum_{cyc}\frac{a}{b+c}\geq\frac{3}{2}.$$ We see that easy to get the expession $2a-b-c$ and it ends the proof by SOS.