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

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 $20$.
