jmc

By QM-AM, $$\sqrt{\frac{x^2+y^2+z^2}{3}}\ge \frac{x+y+z}{3}$$for any nonnegative real numbers $x,$ $y,$ and $z.$ Setting $x=\sqrt{a},$ $y=\sqrt{b},$ $z=\sqrt{c},$ we get $$\sqrt{\frac{a+b+c}{3}}\ge \frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{3}.$$Hence, $$\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{a+b+c}}\le \sqrt{3}.$$Equality occurs when $a=b=c,$ so the largest possible value is $\boxed{\sqrt{3}}.$