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.
 A: I came to my senses -- no need for Calculus.

I see that my solution is similar to Michael Burr's, but anyway, here it is:

Let $f(x,y) = (xy + 2)^2 + (x-1)^2 + (y-1)^2$.

If $x < 0$ then ($x-1)^2 > 1$, so $f(x,y) > 1$.

If $y < 0$ then ($y-1)^2 > 1$, so $f(x,y) > 1$.

If $x,y \ge 0$ then $(xy + 2)^2 \ge 4$, so $f(x,y) > 1$.
A: 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 $\gt 1\,$.
A: If the sum of the three squares is less than or equal to $1$, then each of the squares must, individually, be less than or equal to $1$.  For $(x-1)^2\leq1$, it must be that $0\leq x\leq2$.  Similarly, for $(y-1)^2\leq1$, then $0\leq y\leq2$.  Now, look at $xy+2$, the smallest this could be is $2$, but this leads to a LHS greater than $1$.
A: Because
$$(xy+2)^2+(x-1)^2+(y-1)^2 - 1 = {\frac { \left( x{y}^{2}+x+2\,y-1 \right) ^{2}+3(1+y^2)+ \left( {y}^{2}-y-1 \right) ^{2}}{{y}^{2}+1}}.$$
A: 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 $1$.
