<s>[INST] How do I prove: $xyz(x-2)(y-2)(z-2)\leq\left(1+\frac{2(xy+yz+xz)}{3}\right)^3,\ x,y,z>0$? I have a question which askes to prove:
$\displaystyle \tag*{} xyz(x-2)(y-2)(z-2)\leq\left(1+\frac{2(xy+yz+xz)}{3}\right)^3,\ x,y,z>0$?
My approach: I tried AM-GM inequality, didn't work well. So I tried substituting $2(xy+yz+xz)=(x+y+z)^2-x^2-y^2-z^2$, I couldn't proceed further. I even tried expanding the LHS, it's tedious. Any help would be appreciated, thanks.
 [/INST] We want $x+y+z$ largest when $xyz,xy+yz+zx$ fixed. By 3.83 in the Mathematical Inequalities (Volume 1) by Vasile Cîrtoaje, we can assume $x=y\le z$. Now we rewrite the inequality as $x=y$:
$$ x^2z(x-2)^2(z-2)\leq\left(1+\frac{2(x^2+2xz)}{3}\right)^3$$
We must have $z\ge 2$ otherwise left is smaller than zero, and right is larger than zero.
Now expand everything, we have
$$RHS-LHS=\frac{64x^3z^3+(69x^4+108x^3+36x^2)z^2+(48x^5+54x^4-72x^3+216x^2+108x)z+8x^6+36x^4+54x^2+27}{27}$$
Since $z\ge 2$ so $64x^3z^3\ge72x^3z$. All others are positive, so this is proved.
Remark: I suspect
$$xyz(x-2)(y-2)(z-2)\leq\left(1+\frac{(xy+yz+xz)}{3}\right)^3$$
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