Proving a non-homogeneous inequality with $x,y,z>0$

Question

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 😀 Who can find$?$

Answer 1

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: $$\sum_{cyc}(x^2y^2+x^2)\geq2\sum_{cyc}x^2y\geq\sum_{cyc}(x^2y+xyz).$$