Proving inequality by SOS.

Question

For $x,y,z>0.$ Prove$:$ $$P={x}^{4}y+{x}^{4}z+3\,{x}^{3}{y}^{2}-11\,{x}^{3}yz+3\,{x}^{3}{z}^{2}+3 \,{x}^{2}{y}^{3}+3\,{x}^{2}{y}^{2}z+3\,{x}^{2}y{z}^{2}+3\,{x}^{2}{z}^{ 3}+x{y}^{4}-11\,x{y}^{3}z+3\,x{y}^{2}{z}^{2}\\-11\,xy{z}^{3}+x{z}^{4}+{y }^{4}z+3\,{y}^{3}{z}^{2}+3\,{y}^{2}{z}^{3}+y{z}^{4} \geqq 0$$ There are many SOS way for $P,$ any one can find$?$

For example$,$

NguyenHuyen use fsos function and gave the following expression$:$ And also$:$ $$P=\frac{1}{4} \sum\limits_{cyc} \,z \left( x-y \right) ^{4}+\frac{3}{4} \sum\limits_{cyc} \, \left( {x}^{2}+{y}^{2}+4\,{z}^{2} \right) \left( x-y \right) ^{2}z$$

$(\ast)$ Result by SBM$:$ $$P=\frac{3}{2} \sum\limits_{cyc} \,z \left( xy+2\,{z}^{2} \right) \left( x-y \right) ^{2}+\sum\limits_{cyc} z \left( x-y \right) ^{4}$$

Answer 1

Another way by hand.

By AM-GM twice we obtain: $$\sum_{cyc}(x^4y+x^4z+3x^3y^2+3x^3z^2-11x^3yz+3x^2y^2z)=$$ $$=\sum_{cyc}(x^4y+x^4z-x^3y^2-x^3z^2+4x^3y^2+4x^3z^2-8x^3yz-3x^3yz+3x^2y^2z)=$$ $$=\sum_{cyc}(x-y)^2(xy(x+y)+4z^3-3.5xyz)\geq \sum_{cyc}\left(x-y)^2(2\sqrt{x^3y^3}+4z^3-3.5xyz\right)\geq$$ $$\geq \sum_{cyc}(x-y)^2\left(3\sqrt[3]{\left(\sqrt{x^3y^3}\right)^2\cdot4z^3}-3.5xyz\right)=\left(3\sqrt[3]4-3.5\right)xyz\sum_{cyc}(x-y)^2\geq0.$$

Answer 2

There is also the following SOS: $$\sum_{cyc}(x^4y+x^4z+3x^3y^2+x^3z^2-11x^3yz+3x^2y^2z)=$$ $$=\sum_{cyc}(x^4y+x^4z+3x^3y^2+3x^3z^2-18x^3yz+10x^2y^2z)+\sum_{cyc}7x^3yz-7x^2y^2z)=$$ $$=\sum_{cyc}z(x-y)^2(x+y-3z)^2+7.5xyz\sum_{cyc}(x-y)^2\geq0.$$