jmc

Because we want to find the maximum value of the expression, we can assume that both $x$ and $y$ are positive; if not, then replacing $x$ and $y$ with $|x|$ and $|y|$ would strictly increase the value of the expression.

By Cauchy-Schwarz, $$(1^2+2^2+3^2)(x^2+y^2+1)\ge (x+2y+3{)}^2,$$or $14(x^2+y^2+1)\ge (x+2y+3{)}^2.$ Hence, $$\frac{x+2y+3}{\sqrt{x^2+y^2+1}}\le \sqrt{14}.$$Equality occurs when $x=\frac{y}{2}=\frac{1}{3},$ so the minimum value is $\boxed{\sqrt{14}}.$