Let $x$ and $y$ be nonnegative real numbers. Find the smallest real number $k$ such that

Question

\[\sqrt{x} + \sqrt{y} \le k \sqrt{x + y}\]
for all nonnegative real numbers $x$ and $y.$

Answer 1

We divide both sides by $\sqrt{x + y}$ to get
\[\frac{\sqrt{x}}{\sqrt{x + y}} + \frac{\sqrt{y}}{\sqrt{x + y}} \le k.\]By Cauchy-Schwarz,
\[\frac{\sqrt{x}}{\sqrt{x + y}} + \frac{\sqrt{y}}{\sqrt{x + y}} \le \sqrt{\left( \frac{\sqrt{x}}{\sqrt{x + y}} \right)^2 + \left( \frac{\sqrt{y}}{\sqrt{x + y}} \right)^2} = \sqrt{\frac{x}{x + y} + \frac{y}{x + y}} = 1.\]Therefore, the smallest such $k$ is $\boxed{1}.$