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?
 A: Let $f(x) = \sqrt{x^2+1}+\sqrt{(1-x)^2+1}$. As $x$ and $y$ are non-negative integers, we only consider $x \in [0, 1]$. 
Note that $f(x)$ is symmetric about $x = \frac{1}{2}$ and 
$$f'(x) = \frac{x\sqrt{(1-x)^2+1} + (x - 1)\sqrt{x^2+1}}{\sqrt{x^2+1}\sqrt{(1-x)^2+1}}.$$
For $0 < x < \frac{1}{2}$, $x < 1 - x$ so $\sqrt{x^2+1} > \sqrt{(1-x)^2+1}$. Therefore, $$f'(x) = \frac{x\sqrt{(1-x)^2+1} + (x - 1)\sqrt{x^2+1}}{\sqrt{x^2+1}\sqrt{(1-x)^2+1}} < \frac{\frac{1}{2}\sqrt{(1-x)^2+1} - \frac{1}{2}\sqrt{x^2+1}}{\sqrt{x^2+1}\sqrt{(1-x)^2+1}} < 0.$$ So $f$ is decreasing on $[0, \frac{1}{2})$ and as $f$ is symmetric about $x = \frac{1}{2}$, $f$ is increasing on $(\frac{1}{2}, 1]$. Therefore $f$ attains its minimum value at $x = \frac{1}{2}$ which is $$f\left(\frac{1}{2}\right) = \sqrt{\left(\frac{1}{2}\right)^2+1} + \sqrt{\left(1-\frac{1}{2}\right)^2+1} = \sqrt{\frac{5}{4}} + \sqrt{\frac{5}{4}} = \frac{1}{2}\sqrt{5} + \frac{1}{2}\sqrt{5} = \sqrt{5}.$$ Hence, $f(x) \geq \sqrt{5}$ for $x \in [0, 1]$. Setting $y = 1 -x$ we have $\sqrt{x^2+1} + \sqrt{y^2+1} \geq \sqrt{5}$.
A: If you are familiar with Minkowski's inequality, this is straightforward.
$$\sqrt{x^2+1}+\sqrt{y^2+1} \ge \sqrt{(x+y)^2 + (1+1)^2} = \sqrt5$$
A: Triangle inequality :) The length of the kinked red line represents the sum of the two roots, which is not shorter that the long diagonal of this $1:2$ rectangle.

(Edit: changed my poorly drawn diagram; was this)
A: 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.
A: Another interpretation of the problem is this: $ \ y = \sqrt{x^2 + 1} \ \ \text{and} \ \ x = \sqrt{y^2 + 1} \ $ are the "positive" branches of the "vertical" and "horizontal" unit hyperbolas centered on the origin.  They thus share the asymptotes $ \ y = \pm \ x \ $ . 
It is specified that $ \ x \ge 0 \ \ \text{and} \ \ y \ge 0 \ $ and that they be related through $ \ x + y = 1 \ . $ So we in fact require that $ \ 0 \ \le \ x,y \ \le \ 1 \ . $  We can thus imagine a point in the first quadrant which slides along the line $ \ y = 1 - x \ $ in said quadrant and consider the lengths of the vertical and horizontal line segments extending from each coordinate axis to the appropriate hyperbolic branch, as depicted in the figure below.

It is clear that the behavior of this geometric arrangment is symmetrical about the asymptote $ \ y = x \ , $ so we only need look at the results for, say, $ \ 0 \ \le \ x \ \le \ \frac{1}{2} \ . $  For $ \ x = 0 \ , $ we attain the maximum for the sum sought, 
$$ \sqrt{0^2 + 1} \ + \ \sqrt{1^2 + 1} \ = \ 1 + \sqrt{2} \ \approx \ 2.4142 \ , $$
while for $ \ x = \frac{1}{2} \ , $ the sum has its minimum, 
$$ \sqrt{(\frac{1}{2})^2 + 1} \ + \ \sqrt{(\frac{1}{2})^2 + 1} \ = \ 2 \cdot \sqrt{\frac{5}{4}} \ = \ \sqrt{5} \ \approx \ 2.2361 \ . $$
The sum of radicals has a fairly narrow range of values for the specified domain.  [This approach is most nearly similar to Michael Albanese's analysis.  (It also seems a sort of "inside-out" version of peterwhy's elegant method.)]
A: Construct a right-angled triangle, $1$ unit and $2$ units long for two perpendicular sides. The length of hypotenuse is? And note that straight line gives the shortest distance between two points.
