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
 A: 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, $f(0,3)=3^2-6\cdot 3+11=2.$
A: Your way works very well. The problem is here:
$$\displaystyle k=\bigg[a^2+(b-3)a+\frac{(b-3)^2}{4}+b^2-6b+11-\frac{(b-3)^2}{4}\bigg]$$
when you group the remaining terms in $b$, you lose the $\frac{1}{4}$ factor:
$$\displaystyle k = \bigg[\bigg(a+\frac{b-3}{2}\bigg)^2+\color{red}{\frac{1}{4}}\left(3b^2-18b+35\right)\bigg]$$
With this you arrive at the right result which you probably mean as $2$, not $20$.
A: $$a^2+ab+b^2-3a-6b+11=b^2+(a-6)b+a^2-3a+11=$$
$$=\left(b+\frac{a-6}{2}\right)^2+a^2-3a+11-\frac{(a-6)^2}{4}=$$
$$=\left(b+\frac{a-6}{2}\right)^2+\frac{3a^2+8}{4}\geq2.$$
The equality occurs for $a=0$ and $b=3,$ which says that we got a minimal value.
