Let $x$ and $y$ be nonnegative real numbers. Find the smallest real number $k$ such that
\[\sqrt{x} + \sqrt{y} \le k \sqrt{x + y}\]
for all nonnegative real numbers $x$ and $y.$
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 $\boxed{1}.$
We want to find the smallest $k$ such that $\sqrt{x} + \sqrt{y} \le k \sqrt{x + y}$ for all nonnegative $x$ and $y.$ Squaring both sides, we get
\[x + 2 \sqrt{xy} + y \le k^2 (x + y).\]We have $\sqrt{xy} \le \frac{x + y}{2},$ so
\[x + 2 \sqrt{xy} + y \le x + x + y + y = 2(x + y).\]Hence,
\[2(x + y) \le k^2 (x + y)\]for all nonnegative $x$ and $y.$ This reduces to
\[(2 - k^2)(x + y) \le 0,\]which holds if and only if $k^2 \le 2.$ The smallest such $k$ is $\boxed{\sqrt{2}}.$
To find the smallest real number $k,$ we want to find the largest possible value of $\frac{\sqrt{x} + \sqrt{y}}{\sqrt{x + y}}.$ In other words, we want to maximize this expression.
To do this, we can apply the Cauchy-Schwarz inequality. This states that for all real numbers $a_1,$ $a_2,$ $b_1,$ $b_2,$ $\dots,$ $b_n,$
\[(a_1^2 + a_2^2 + \dots + a_n^2)(b_1^2 + b_2^2 + \dots + b_n^2) \ge (a_1 b_1 + a_2 b_2 + \dots + a_n b_n)^2.\]In this case, we take $n = 2,$ $a_1 = \sqrt{x},$ $a_2 = \sqrt{y},$ $b_1 = \frac{1}{\sqrt{x}},$ and $b_2 = \frac{1}{\sqrt{y}}.$ Then the inequality becomes
\[(x + y)\left( \frac{1}{x} + \frac{1}{y} \right) \ge (\sqrt{x} \cdot \frac{1}{\sqrt{x}} + \sqrt{y} \cdot \frac{1}{\sqrt{y}})^2,\]which simplifies to
\[\left( 1 + \frac{y}{x} + \frac{x}{y} \right) \ge (\sqrt{x} \cdot \frac{1}{\sqrt{x}} + \sqrt{y} \cdot \frac{1}{\sqrt{y}})^2 = (\sqrt{x} + \sqrt{y})^2.\]Hence,
\[\frac{\sqrt{x} + \sqrt{y}}{\sqrt{x + y}} \le \sqrt{1 + \frac{y}{x} + \frac{x}{y}}.\]We want to find the largest possible value of $\sqrt{1 + \frac{y}{x} + \frac{x}{y}}.$ By AM-GM or other means, we can show that
\[\frac{y}{x} + \frac{x}{y} \ge 2,\]so
\[1 + \frac{y}{x} + \frac{x}{y} \ge 3.\]Therefore,
\[\frac{\sqrt{x} + \sqrt{y}}{\sqrt{x + y}} \le \sqrt{3}.\]Equality occurs when $\frac{x}{y} = 1,$ so the smallest possible value of $k$ is $\boxed{\sqrt{3}}.$