<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] 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 $\sqrt{x^2+1} + \sqrt{y^2+1} \geq \sqrt{5}$.
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