# The smallest value of the expression $4x^2y^2+x^2+y^2-2xy+x+y+1$

What is the smallest value that $$4x^2y^2+x^2+y^2-2xy+x+y+1$$ can take with real numbers $$x$$ and $$y$$?

I suspect the following transformation can be done: $$(2xy-1/2)^2 + (x+1/2)^2 + (y+1/2)^2 + 1/4$$.

• Then, you need to take $x=y=-\frac 12$. I can not see another problem. Because, if $x=y=-\frac 12$, then all sum of of square equals to zero. Otherwise we would need another technique. Note that, all quadratic expressions can be zero at the same time. Oct 16 at 19:59
• Your "completed square" version is right. So that shows it's always at least $1/4.$ But the squared things aren't independent of each other so something more is needed. Of course if one could find a pair $(x,y)$ realizing $1/4$ that would be the min. Note using @lonestudent suggestion we do get $1/4.$ Oct 16 at 19:59
• @coffeemath Sure, for instance,we can take $$f(x,y)=\left(2xy-\frac 15\right)^2+\left(x+\frac 12\right)^2+\left(y+\frac 12\right)^2+\frac 14$$ So, $x=y=-\frac 12$ doesn't work. Oct 16 at 20:54
• @lonestudent I was just referring to OP's speecific function. But as you just noted one can't just put something into a sum of squares plus a constant and conclude that constant is the min. I wonder if there's a general method to get the min of any quadratic function of two vars... Oct 16 at 21:16
• @coffeemath Yes, I was just supporting your comment. It is really necessary to check if the first expression is zero. Maybe it is possible to reach a general conclusion by applying the same method. But in our case, the degree of the polynomial $4$. So our job is a little difficult. I didn't try. It can work or not. But, it worked for degree $2$ polynomial. Oct 16 at 21:37

$$(2xy−\frac12)^2+(x+\frac12)^2+(y+\frac12)^2+\frac14 \geq \frac14$$.

Because $$(2xy−\frac12)^2\geq 0,(x+\frac12)^2 \geq 0,(y+\frac12)^2 \geq 0$$. So this minimum is attained in the original expression when $$(2xy−\frac12)^2=0,(x+\frac12)^2=0,(y+\frac12)^2=0 \iff x=y=-\frac12$$.

After $$(2xy-1/2)^2 + (x+1/2)^2 + (y+1/2)^2 + 1/4$$, use a change of variables $$x - 1/2 = u, y - 1/2 = v$$, and $$(2xy-1/2) = 2(u + 1/2)(v + 1/2) - 1/2$$ $$= 2(uv + u/2 + v/2 + 1/4) - 1/2$$ to get:

$$(2uv+u+v)^2 + u^2 + v^2 + 1/4$$ $$=(2uv+u+v)^2 + (u+v)^2 - 2uv + 1/4$$ $$= (p+q)^2 + q^2 - p + 1/4$$

where $$p=2uv, q = u+v$$.

For $$u, v$$ to be real numbers, $$q^2 - p ≥ 0$$ as $$u^2+v^2 ≥ 0$$ for all real numbers $$u,v$$. Thus $$(p+q)^2 + q^2 - p + 1/4 ≥ (p + q)^2 + 1/4 ≥ 1/4$$.

This minimum is attained in the original expression when $$x = y = -1/2$$.