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?
Solution 1:
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)
Solution 2:
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.
Solution 3:
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 $\sqrt{x^2+1} + \sqrt{y^2+1} \geq \sqrt{5}$.
Solution 4:
If you are familiar with Minkowski's inequality, this is straightforward.
$$\sqrt{x^2+1}+\sqrt{y^2+1} \ge \sqrt{(x+y)^2 + (1+1)^2} = \sqrt5$$