# Prove $\sqrt{x^2+1} + \frac{1}{\sqrt{x^2 +1}}\geq 2$ [duplicate]

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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 $x^2>0$

## marked as duplicate by Martin R, Isaac Browne, Strants, Batominovski, José Carlos SantosJul 30 '18 at 22:36

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• Hint: think AM-GM, or any of several related ways. – dxiv Jul 30 '18 at 6:42
• I don't see how to use AM-GM here? – john fowles Jul 30 '18 at 6:48
• By AM-GM $\;\displaystyle \frac{\;a + \dfrac{1}{a}\;}{2} \ge \sqrt{a \cdot \frac{1}{a}} = 1\,$, then use it for $\,a=\sqrt{x^2+1}\,$. – dxiv Jul 30 '18 at 6:51
• This is very nice! – john fowles Jul 30 '18 at 6:53
• – Martin R Jul 30 '18 at 8:10

## 2 Answers

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

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

• f increases in [1,∞),And also decreases on $[0,1]$ which is needed to complete the argument. – dxiv Jul 30 '18 at 6:45
• @dxiv - Since $\sqrt{x^2+1}\geq 1$ for all $x\in\mathbb{R}$, the interval $[0,1]$ is irrelevant. – uniquesolution Jul 30 '18 at 6:51
• Right, but that belongs in the answer, not a comment. – dxiv Jul 30 '18 at 6:53
• Feel free to include this in your answer. – uniquesolution Jul 30 '18 at 6:53
• I didn't post an answer. Feel free to spell it out in your answer. – dxiv Jul 30 '18 at 6:54