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Let $x,y,z\ge 0$,$x+y+z=3$,prove: $$3{{x}^{2}}(1+2y){{(1+z)}^{3}}+3{{y}^{2}}(1+2z){{(1+x)}^{3}}+3{{z}^{2}}(1+2x){{(1+y)}^{3}}\le {{\left( 3+xy+yz+zx \right)}^{3}}$$

my idea:let $p=x+y+z=3,q=xy+yz+xz,r=xyz$

then $$LHS=6xyz(x^2y+y^2z+z^2x)+3(x^3y^2+y^3z^2+z^3x^2)+18xyz(xy+yz+xz)+9(x^2y^2+y^2z^2+x^2z^2)+18xyz(x+y+z)+6(x^2y+y^2z+z^2x)+9(x^2z+y^2x+z^2y)$$

and so $LHS=f(p,q,r)$,can see:How prove this $(abc)^4+abc(a^3c^2+b^3a^2+c^3b^2)\le 4$ follow is very ugly.

or this equality has other methods,Thank you everyone.

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