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 $$\sum_{cyc}\frac{a^3}{bc}\ge a+b+c$$
 A: 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.
A: 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
$$\sum \frac{(a-b)^2[(a+b)^2+c^2]}{2} \geqslant 0.$$
A: 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*}
A: 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, $$F\ge (a+b+c)$$
A: There are many SOS!
My SOS, first is same as Mr. Mike$:$
\begin{align*} \sum \frac{a^3}{bc} -\sum a &=\dfrac{1}{2} \sum {\dfrac { \left( {a}^{2}+ab+ac+{b}^{2}+bc \right)  \left( a-b
 \right) ^{2}}{bca}}\\&=\dfrac{1}{4}\sum {\frac { \left( 3\,a+4\,b \right)  \left( a-b \right) ^{2}}{bc}}+
\dfrac{1}{4}\sum{\frac {a \left( a+b-2\,c \right) ^{2}}{bc}}\\&
=\sum{\dfrac { \left( a+b \right)  \left( a-b \right) ^{2}}{ab}}+\dfrac{1}{6}\sum{
\dfrac { \left( 2\,{a}^{2}-{b}^{2}-{c}^{2}+2\,bc-ab-ac \right) ^{2}}{bc
a}}
\\&=\dfrac{1}{7}\sum {\dfrac { \left( a-b \right) ^{2} \left( 5\,a+8\,c \right) }{ac}}+ \dfrac{2}{7}\sum {\dfrac { \left( {a}^{2}-2\,ab+bc \right) ^{2}}{bca}}+\\&\quad 
+{\frac {5}{ 42}}\sum{\frac { \left( 2\,{a}^{2}-{b}^{2}-{c}^{2}+2\,bc-ab-ac \right) ^ {2}}{bca}}
\end{align*}
