<s>[INST] How to prove that $(xy+2)^2+(x-1)^2+(y-1)^2 > 1$ for real $x,y$ How do I prove that $$(xy+2)^2+(x-1)^2+(y-1)^2 > 1$$ for all real $x,y$? I tried setting the partial derivative to $0$ but I couldn't get it to work.
 [/INST] Let $\,p=xy\,$ and $\,s=x+y\,$, then the LHS is:
$$
\begin{align}
(xy+2)^2 + (x-1)^2+(y-1)^2 & = (p+2)^2 + (x^2+y^2) - 2(x+y) + 2 \\
 & = (p^2 + 4p + 4)+(s^2-2p) -2s + 2 \\
 & = p^2+2p+s^2-2s+6 \\
 & = (p+1)^2 +(s-1)^2+ 4 \\
 & \ge 4 
\end{align}
$$
Equality is attained for $p=-1,s=1$ i.e. when $x,y$ are the roots of $t^2-t-1=0\,$.
Of course, the LHS being $\ge 4$ implies that it's $\gt 1\,$.
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