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This question already has an answer here:

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

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

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

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

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