<s>[INST] Minimum value of $f(a,b) = a^2+ab+b^2-3a-6b+11$ 
Minimum value of
$f(a,b) = a^2+ab+b^2-3a-6b+11$ for all $a,b\in \mathbb{R}$

what i try
Let $$k=a^2+ab+b^2-3a-6b+11$$
$$k=a^2+(b-3)a+b^2-6b+11$$
$\displaystyle k=\bigg[a^2+(b-3)a+\frac{(b-3)^2}{4}+b^2-6b+11-\frac{(b-3)^2}{4}\bigg]$
$$\displaystyle k = \bigg[\bigg(a+\frac{b-3}{2}\bigg)^2+3b^2-18b+35\bigg]$$
$$k=\bigg[\bigg(a+\frac{b-3}{2}\bigg)^2+3(b-3)^2+8\bigg]\geq 8$$
but answer is $20$
Help me please
 [/INST] Note that as $a \to \pm \infty$ or $b\to \pm \infty$, the function $f$ diverges positively so the boundary will not lead to the infinimum. Using the standard technique,
\begin{align*}
f_a=2a+b-3=0\implies b=3-2a
\end{align*}
\begin{align*}
f_b=2b+a-6=0\implies a=6-2b.
\end{align*}
Hence, the critical point occurs when $b=3-2(6-2b)=4b-9\implies b=3.$ Then it easily follows that $a=0$. Plugging in, $f(0,3)=3^2-6\cdot 3+11=2.$
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