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.

Hint (1): Why did you choose the case $2ab = -1$ ? Well, as you showed, this is the smallest possible value for $2ab$ . What can you say about $|a+b|$ when $2ab > -1$ ?
Hint (2) for a different method: Consider a circle of radius 1 around the center of coordinates. The equation of this circle is $x^2 + y^2 = 1$ . Now consider the set of parallel lines $x+y = c$ where $c$ can be any number. Some of these lines touch the circle. Among those lines, which one has the highest value of $c$ ?
