<s>[INST] least value of $f(x,y) = 2x^2+y^2+2xy+2x-3y+8\;,$ Where $x,y\in \mathbb{R}$ 
The least value of $f(x,y) = 2x^2+y^2+2xy+2x-3y+8\;,$ Where $x,y\in \mathbb{R}$

$\bf{My\; Try::}$ Let $$K = 2x^2+y^2+2xy+2x-3y+8$$
So $$\displaystyle y^2+(2x-3)y+2x^2+2x+8-K=0$$
Now For real values of $y\;,$ We have $\bf{Discriminant\geq 0}$
So $$(2x-3)^2-4(2x^2+2x+8-K)\geq 0$$
So $$-4x^2-20x-23+4K\geq0\Rightarrow 4x^2+20x+23-4K\leq0$$
Now How can I Solve after that, Help me
Thanks 
 [/INST] \begin{align}
&2x^2+y^2+2xy+2x-3y+8\\
&=\left(x^2+y^2+\frac94+2xy-3x-3y\right)+x^2+5x+\frac{23}{4}\\
&=(x+y-3/2)^2+x^2+5x+\frac{23}4\\
&=(x+y-3/2)^2+\left(x+\frac52\right)^2-\frac12\\
&\geq-\frac12\\
&(x=-\frac52,\space y=4)
\end{align}
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