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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: $(2xy-1/2)^2 + (x+1/2)^2 + (y+1/2)^2 + 1/4$.

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    $\begingroup$ 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. $\endgroup$ Oct 16 at 19:59
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    $\begingroup$ 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 $1/4.$ $\endgroup$
    – coffeemath
    Oct 16 at 19:59
  • $\begingroup$ @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. $\endgroup$ Oct 16 at 20:54
  • $\begingroup$ @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... $\endgroup$
    – coffeemath
    Oct 16 at 21:16
  • $\begingroup$ @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. $\endgroup$ Oct 16 at 21:37
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$(2xy−\frac12)^2+(x+\frac12)^2+(y+\frac12)^2+\frac14 \geq \frac14$.

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 $(2xy−\frac12)^2=0,(x+\frac12)^2=0,(y+\frac12)^2=0 \iff x=y=-\frac12$.

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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 $.

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