Prove $\sqrt{x^2+1} + \frac{1}{\sqrt{x^2 +1}}\geq 2$

Question

I want to show why the last inequality in the problem below $\sqrt{x^2+1} + \frac{1}{\sqrt{x^2 +1}}\geq 2$ holds. It's clear that $x^2\geq 0$ and that equality holds when $x=0$ but how can I clearly show this. The left hand side, for $x^2>0$ is greater than $1$ but the right hand side becomes less than $1$ as $x^2>0$

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Answer 1

The inequality $(y-1)^{2} \geq 0$ gives $y+\frac 1 y \geq 2$ for any positive number $y$. Take $y=\sqrt {1+x^{2}}$.

Answer 2

The derivative of $f(y)=y+1/y$ is $1-1/y^2$, so the function $f$ increases in $[1,\infty)$, hence $f(y)\geq f(1)=2$.