if $a^{2}+b^{2}=1$ Prove that $-\sqrt{2} \le a+b \le \sqrt{2}$ $a,b \in \mathbb{R}$
My attempt:
$$a^{2}+b^{2}=1 \iff a^{2}+2ab+b^{2}=1+2ab$$
$$(a+b)^2=1+2ab \iff \mid a+b \mid =\sqrt{1+2ab} \iff a+b = \pm\sqrt{1+2ab}$$
But notice that $\sqrt{1+2ab} \in \mathbb{R} \iff 1+2ab \geq 0 \iff 2ab \geq -1$ .
let's take the case where :$2ab=-1$.
$$a^{2}+2ab+b^{2}=0 \iff a+b=0 \iff a=-b.$$
When I plug this result into the original equation, I get:
$$b=\frac{1}{\sqrt{2}}$$
And now, I don't know where to go.

Using the triangle inequality and the equivalence between the $\ell^1$ and the $\ell^2$ norm, we have
$$|a+b| \leq |a|+|b| = \|(a,b)\|_1 \leq \sqrt{2}\, \|(a,b)\|_2 = \sqrt{2}\,\sqrt{a^2+b^2} = \sqrt{2}.$$
