If $x>0$ how can I show that $\frac{1}{2\sqrt{1+x}}<\sqrt{x+1}-\sqrt{x}<\frac{1}{2\sqrt{x}}$?

I tried proving this by proving the inequalities seperately and trying to let one side equal $f(x)$ and the other $g(x)$ and therefore using $h(x)=f(x)-g(x)$ to try and show $h(x)$ is either nonnegative or nonpositive, but I could not really get through it and I do not really know if that is the approach for this one, would appreciate the help.


I think you're supposed to see $\sqrt{x + 1} - \sqrt{x}$ as $$\frac{\sqrt{x + 1} - \sqrt{x}}{(x + 1) - x}.$$ MVT implies that there is some $y$ between $x$ and $x + 1$ such that $$\frac{1}{2\sqrt{y}} = \frac{\sqrt{x + 1} - \sqrt{x}}{(x + 1) - x} = \sqrt{x + 1} - \sqrt{x}.$$ The rest is using the fact that $\frac{1}{2\sqrt{y}}$ is a strictly decreasing function.

  • $\begingroup$ I was able to work out to where you arrived but not sure I understand how to relate 1/2sqrt(y) being a strictly decreasing function with the rest. Edit: just as I posted this I figured it out, thank you! $\endgroup$ Apr 2 '20 at 3:53
  • 1
    $\begingroup$ Well, you know that $x < y < x + 1$, from the conclusion of the MVT. So, since $\frac{1}{2\sqrt{y}}$ is strictly decreasing, this implies $\frac{1}{2\sqrt{x}} > \frac{1}{2\sqrt{y}} > \frac{1}{2\sqrt{x + 1}}$, as necessary. $\endgroup$
    – user764828
    Apr 2 '20 at 3:55
  • $\begingroup$ Thank you, I was able to figure it out seconds after I added my comment lol! $\endgroup$ Apr 2 '20 at 3:57


Start by noticing that

\begin{align*} \sqrt{x+1} - \sqrt{x} = (\sqrt{x+1} - \sqrt{x})\times\frac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1} + \sqrt{x}} = \frac{1}{\sqrt{x+1}+\sqrt{x}} \end{align*}

Then observe that $\sqrt{x} < \sqrt{x+1}$, from whence one gets \begin{align*} 2\sqrt{x} < \sqrt{x} + \sqrt{x+1} < 2\sqrt{x+1} \end{align*}


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