Inequality involing square-root

Question

I want to show for $x>0$ the inequality $$ \sqrt{x+1}-\sqrt{x} \leq \frac{1}{\sqrt{x}} , $$ which gets suprisingly sharp for large values of $x$.

I guess some kind of binomial theorem for non-integer exponents would be helpful. But maybe there is something simpler?

Any hints? thanks.

Answer 1

This bound is not the sharpest you can think of. In fact,

$$\sqrt{x+1}-\sqrt x\le\frac1{\color{green}2\sqrt x}.$$ (This can be shown using Michael Rozenberg's approach.)

There are several ways to obtain this relation:

1. let $h=1$. Then for large $x$, $\dfrac{\sqrt{x+h}-\sqrt x}h$ is an approximation of $\left(\sqrt x\right)'=\dfrac1{2\sqrt x}$.

2. $\sqrt{x+1}-\sqrt x=\sqrt x\left(\sqrt{1+\dfrac1x}-1\right)\approx \sqrt x\left(1+\dfrac1{2x}-1\right)=\dfrac1{2\sqrt x}$ by the generalized binomial theorem or by Taylor.

Answer 2

It's $$\frac{(\sqrt{x+1}-\sqrt{x})(\sqrt{x+1}+\sqrt{x})}{\sqrt{x+1}+\sqrt{x}}\leq\frac{1}{\sqrt{x}}$$ or $$\frac{1}{\sqrt{x+1}+\sqrt{x}}\leq\frac{1}{\sqrt{x}},$$ which is obvious.

Answer 3

multiplying numerator and denominator of the left side by $$\sqrt{x+1}+\sqrt{x}$$ and you will get $$\frac{1}{\sqrt{x+1}+\sqrt{x}}\le \frac{1}{\sqrt{x}}$$

Answer 4

Here's another way to see it. The inequality you give is equivalent to

$$ 1 \le \frac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x}} = 1+ \frac{\sqrt{x+1}}{\sqrt{x}}. $$

This way of looking at it also makes it relatively transparent why you should expect the inequality to be sharp for large $x$.