Inequality with square roots: $\sqrt{x^2+1}+\sqrt{y^2+1}\ge \sqrt{5}$ 
Let $x$ and $y$ be nonnegative real numbers such that $x+y=1$. How do I show that $\sqrt{x^2+1}+\sqrt{y^2+1}\ge \sqrt{5}$?

How do I deal with square roots inside the inequality?

Assuming the contrary, for some $0\le x \le 1$, and $y=1-x$,
$$\begin{align}
\sqrt{x^2+1}+\sqrt{y^2+1} <& \sqrt{5}\\
\left( \sqrt{x^2+1} + \sqrt{y^2+1} \right)^2 <& 5 &\text{(Square both sides)}\\
x^2 +1 + 2\sqrt{ \left( x^2 + 1 \right)\left( y^2 + 1 \right)} + y^2 + 1 <& 5\\
2\sqrt{ \left( x^2 + 1 \right)\left( y^2 + 1 \right)} <& 3 - x^2 -y^2\\
2\sqrt{ \left( x^2 + 1 \right)\left( y^2 + 1 \right)} <& 3 - \left(x+y\right)^2 + 2xy\\
2\sqrt{ \left( x^2 + 1 \right)\left( y^2 + 1 \right)} <& 3 - 1 + 2xy &\text{(Substitute }x+y=1\text{)}\\
\sqrt{\left( x^2+1 \right) \left( y^2 + 1 \right)} <& 1+xy\\
\left( x^2+1 \right) \left( y^2 + 1 \right) <& \left( 1+xy \right)^2 &\text{(Square both sides)}\\
x^2 y^2 + x^2 + y^2 + 1 <& 1 + 2xy + x^2 y^2\\
x^2 -2xy+y^2 <& 0\\
\left( x - y \right)^2 <& 0\\
\end{align}$$
which contradicts.
