<s>[INST] Inequality Proof $(x^2+1)(y^2+1)(z^2+1)\leq...$ I want to show, that for positive numbers $x,y,z$ with $xy,yz,zx\geq1$, $(x^2+1)(y^2+1)(z^2+1)\leq\left(\left(\frac{x+y+z}{3}\right)^2+1\right)^3$.
Using the AM-QM inequality for $(x^2+1),(y^2+1)$ and $(z^2+1)$, I already got $(x^2+1)(y^2+1)(z^2+1)\leq\left(\frac{x^2+y^2+z^2}{3}+1\right)^3$.By equivalent transformations of $\left(\frac{x^2+y^2+z^2}{3}+1\right)^3\leq\left(\left(\frac{x+y+z}{3}\right)^2+1\right)^3$ I get $0\leq\frac{2}{3}\left(xy+yz+zx-x^2-y^2-z^2\right)$, which cannot be true. Where did I miss something? Is the AM-QM inequality too weak? 
Thank you!
 [/INST] Hint: First show that, if $ab\geq 1$,
$$(a^2+1)(b^2+1)\leq \left(\left(\frac{a+b}{2}\right)^2+1\right)^2$$
(hint for this: expand out and use that $(a-b)^2\geq 0$ and $ab\geq 1$). Then apply this to get that
$$(a^2+1)(b^2+1)(c^2+1)(d^2+1)\leq \left(\left(\frac{a+b+c+d}{4}\right)^2+1\right)^4$$
if $ab,bc,cd,da,ac,bd\geq 1$. Now, set $d=\frac{a+b+c}{3}$.
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