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.

Using @quasi 's original comment, consider the function
$$
f(x,y)=(xy+2)^2+(x-1)^2+(y-1)^2.
$$
If we take partial derivatives of $f$, we get
\begin{align*}
\frac{\partial f}{\partial x}&=2(xy+2)y+2(x-1)\\
\frac{\partial f}{\partial y}&=2(xy+2)x+2(y-1).
\end{align*}
At minima, these are equal to zero, so we consider
\begin{align*}
2(xy+2)y+2(x-1)&=0\\
2(xy+2)x+2(y-1)&=0.
\end{align*}
Next, we can subtract the second equation from the first equation to get
$$
2(xy+2)(y-x)+2(x-y)=0.
$$
Factoring this, we have
$$
2(y-x)(xy+2-1)=0
$$
or that
$$
2(y-x)(xy+1)=0.
$$
Therefore, minima occur when, either
1. $x=y$ or
2. $xy+1=0$.
In the first case, $(xy+2)^2$ simplifies to $(x^2+2)^2$, since squares are always positive, the inside of the square is at least $2$, so the LHS is at least $4$.
In the second case, if $xy+1=0$, then $xy=-1$, and, plugging this into the formula for $f$, we see that the LHS is at least $(xy+2)^2=(-1+2)^2=1$.  Now, we can't have both $x=1$ and $y=1$ and have their product $-1$, so at least one of $(x-1)^2$ or $(y-1)^2$ is positive, and the LHS is greater than $1$.
