I recall that the square root function is concave, which means that it satisfies the inequality $\sqrt{\frac{x + y}{2}} \geq \frac{\sqrt{x} + \sqrt{y}}{2}$ for any nonnegative $x$ and $y$. This inequality can be extended to more than two variables by repeated application, so I have $\sqrt{\frac{a + b + c}{3}} \geq \frac{\sqrt{a} + \sqrt{b} + \sqrt{c}}{3}$. Multiplying both sides by $\sqrt{3}$, I get $\sqrt{a + b + c} \geq \frac{\sqrt{a} + \sqrt{b} + \sqrt{c}}{\sqrt{3}}$. Taking the reciprocal of both sides, I get $\frac{1}{\sqrt{a + b + c}} \leq \frac{\sqrt{3}}{\sqrt{a} + \sqrt{b} + \sqrt{c}}$. Multiplying both sides by $\sqrt{a} + \sqrt{b} + \sqrt{c}$, I get $\frac{\sqrt{a} + \sqrt{b} + \sqrt{c}}{\sqrt{a + b + c}} \leq \sqrt{3}$. Therefore, the largest possible value of the expression is $\sqrt{3}$, and it is achieved when $a = b = c$. 
Let $a,$ $b,$ $c$ be positive real numbers.  Find the largest possible value of
\[\frac{\sqrt{a} + \sqrt{b} + \sqrt{c}}{\sqrt{a + b + c}}.\]