<s>[INST] 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?
 [/INST] 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.)]
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