Find $a^2+b^2+2(a+b)$ minimum if $ab=2$ Let $a,b\in R$,and such 
$$ab=2$$
Find the minimum of the $a^2+b^2+2(a+b)$.
I have used $a=\dfrac{2}{b}$, then
$$a^2+b^2+2(a+b)=\dfrac{4}{b^2}+b^2+\dfrac{4}{b}+2b=f'(b)$$
Let $$f'(b)=0,\,b=-\sqrt{2}$$
So $$a^2+b^2+2(a+b)\ge 4-4\sqrt{2}$$
I wanted to know if there is other way to simplify the function and find the required value without using messy methods. Can we cleanly use AM-GM inequality?

For $a=b=-\sqrt2$ we get a value $4-4\sqrt2$.
We'll prove that it's a minimal value.
Indeed, let $a+b=2k\sqrt{ab}$. 
Hence, $|k|=\left|\frac{a+b}{2\sqrt{ab}}\right|\geq1$ and we need to prove that
$$a^2+b^2+2(a+b)\geq4-4\sqrt2$$ or
$$a^2+b^2+\sqrt{2ab}(a+b)\geq(2-2\sqrt2)ab$$ or
$$(a+b)^2+\sqrt{2ab}(a+b)\geq(4-2\sqrt2)ab$$ or
$$4k^2+2\sqrt2k\geq4-2\sqrt2$$ or
$$(k+1)(2k-2+\sqrt2)\geq0,$$
which is obvious for $|k|\geq1$.
Done!
