Having a bit of trouble with this one. Can anyone help?

Many thanks.

Prove $\displaystyle a^2+2ab+2b^2\geq0$

Q.

Attempt:My original solution was going to be...: if $\displaystyle (a+b)(a+2b)\geq0$

However, this does not fit the format of (real no.)^{2}>0

The alternative method I've found goes as follows: $\displaystyle a^2+2ab+2b^2\geq0$

if $\displaystyle a^2+4ab+4b^2\geq0$

if $\displaystyle (a+b)^2$...true, since (a+b)^{2}>0

Is this 2nd approach correct? How did they go from $\displaystyle a^2+2ab+2b^2\geq0$ to $\displaystyle a^2+4ab+4b^2\geq0$?