After reading many interesting approaches to this question, most of them using Calculus, I decided to try to offer an elementary solution.
If [math] a ~& ~b [/math] are such that [math]a^2 + b^2 = 1,[/math] we can estimate:
[math]0\le (b-a)^2 = b^2+a^2 -2ab = 1-2ab \implies 2ab \le 1.[/math]
Hence,
math^2 = a^2 + b^2 +2ab = 1+2ab \le 1 + 1 = 2.[/math]
In conclusion:
[math]a^2+b^2 = 1 \implies -\sqrt{2} \le a+b \le \sqrt{2}.[/math]
The upper bound is attained at [math]a=b=\frac{\sqrt{2}}{2}[/math] and the lower bound is attained at [math]a=b=-\frac{\sqrt{2}}{2}[/math].
Thus, the maximum of the sum is [math]\sqrt{2}[/math] and the minimum [math]-\sqrt{2}.[/math]
Ps.: As a byproduct, we have also shown that if [math]a^2 + b^2 = 1[/math] then [math]-\frac{1}{2}\le ab \le \frac{1}{2}.[/math] The extrema are also attained when [math]a=b=\pm \frac{\sqrt{2}}{2}.[/math]
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