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 $$\sum_{cyc}\frac{a^3}{bc}\ge a+b+c$$

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*}
