Solve $\sqrt{\left(a+1\right)^2+b^2}\ge2+a$

Question

I stumbled across this inequation and I am stuck at the solution: $$\tag{$\forall a,b \in \mathbb{R}$} \sqrt{\left(a+1\right)^2+b^2}\ge2+a$$

It is equal to: $$ \tag{if a ≥ −2} =\left(a+1\right)^2+b^2\ge\left(2+a\right)^2\ \ \ $$

$$a \ge (b^2-3)\cdot\frac{1}{2}$$

But I am not able to get rid of the root when $a <-2$: $$ =\sqrt{\left(a+1\right)^2+b^2}\ge2+a $$ In the first part I could just square both sides because both sides were non-negative. But this is not the case if $a <-2$ How could this be solved?

Answer 1

When $a <-2$, RHS is negative, whereas LHS is always positive. Hence this is always true. Therefore the solution will be

$$b \in \Bbb R , a \in (-\infty,-2)$$

Answer 2

If $a<-2$, then the right hand side is $a+2<0$, but notice that the left hand side is always nonnegative.

Hence $a<-2$ would satisfy the inequality regardless of the values of $b$.