OlympiadHardInequalities10–12

For positive reals, if $a^{2} + b^{2} = 1$ , what is the maximum value of $a + b$ ?

Para reais positivos, se $a^{2} + b^{2} = 1$ , qual é o valor máximo de $a + b$ ?

Answer options

Solution

By Cauchy-Schwarz: $(a+b)^{2} \le 2(a^{2}+b^{2}) = 2$ , so $a+b \le \sqrt{2}$ . Maximum when $a=b=\frac{1}{\sqrt{2}}$ .