<s>[INST] Find the smallest possible value for $x^2-3x+2y^2+4y+2$ Find the smallest possible value for $x^2-3x+2y^2+4y+2$.
I know this: $x^2-3x+2y^2+4y+2=(x-\frac{3}{2})^2+2(y+1)^2-\frac{9}{4}$ so the smallest value is $-\frac{9}{4}$.
Now we also know this: $x^2-3x+2y^2+4y+2=(x-1)^2-1-x+2(y+1)^2=(*)(x-1)^2+2(y+1)^2-(1+x)$
My question is: why we cannot say that the smallest value possible is $-(1+x)$ when $(x-1)^2=0$ (i.e. when $x=1$) and $(y+1)^2=0$, and because $x=1$ so the smallest value is $-2$?
 [/INST] $x^2-3x+2y^2+4y+2=0$ is an ellipse, so what do you mean by "the smallest possible value of $x^2-3x+2y^2+4y+2$"?
Anyway, since you have two distinct values of $y=y(x)$ for the given $x: x=\frac{3}{2}$ such that one of them (the negative one) is minimum, I will write only the positive value of $y(x) : x=\frac{3}{2}$, leaving the final step (and its proof) as an exercise.
Well, $y\left(\frac{3}{2}\right) : y>0$ is equal to $\left(\frac{3}{2\cdot \sqrt{2}} - 1\right)$ so that you have to show why $y\left(\frac{3}{2}\right) : y<0$ is the minimum and then find the value.
P.S. If we are just interested in the global minimum of $x^2-3x+2y^2+4y+2$ in $\mathbb{R}^2$, it is at $(x,y)=(\frac{3}{2},-1)$ and it is $-\frac{9}{4}$ indeed.
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