Proving a non-homogeneous inequality with $x,y,z>0$ For $x,y,z>0.$ Prove: $$\frac{1}{2}+\frac{1}{2}{r}^{2}+\frac{1}{3}\,{p}^{2}+\frac{2}{3}\,{q}^{2}-\frac{1}{6} Q-\frac{3}{2} r-\frac{2}{3}q-\frac{1}{6}pq-\frac{5}{3} \,pr\geqslant 0$$
where $$\Big[p=x+y+z,q=xy+zx+yz,r=xyz,Q= \left( x-y \right)  \left( y-z \right) 
 \left( z-x \right)\Big ]$$
My SOS proof$:$ $$\text{LHS}=\frac{1}{12}\,\sum \left( 3\,{z}^{2}+1 \right)  \left( x-y \right) ^{2}+\frac{1}{6} \sum\,y
 \left( y+z \right)  \left( x-1 \right) ^{2}+\frac{1}{2}\, \left( xyz-1
 \right) ^{2} \geqslant 0$$
By the way$,$ there is an C-S proof :D Who can find$?$

Since by AM-GM $$1+x^2y^2z^2\geq2xyz$$ and $$\sum_{cyc}x^2y^2\geq\sum_{cyc}x^2yz,$$ it's enough to prove that:
$$\sum_{cyc}x^2y^2-\sum_{cyc}(x^2y+xyz)+\sum_{cyc}x^2\geq0,$$ which is true by AM-GM twice:
$$\sum_{cyc}(x^2y^2+x^2)\geq2\sum_{cyc}x^2y\geq\sum_{cyc}(x^2y+xyz).$$
