Why is $2\sqrt{x+1}-2\geq\frac{\sqrt{x}}{2}$ for $x \in \mathbb{N}$? I have found following formula stands:
$2\sqrt{x+1}-2\geq\frac{\sqrt{x}}{2}$
($x \in \mathbb{N}$)
I have wondered why this is true. I have checked its rightness on the graph drawing tools, but I want to know the mathematical proof of this.
Although I can prove it for $x \in \mathbb{N}, x\geq2$, as following:
$2\sqrt{x+1} - \sqrt{x}/2 \geq 2\sqrt{x} - \sqrt{x}/2 = \frac{3}{2}\sqrt{x}\geq\frac{3}{2}\cdot \sqrt{2} \geq2$
and prove the rightness in case $x=1$ by direct calculation,
I want to know whether it is possible to prove this formula without dividing the case of $x=1$ and others.
Thank you.

Using the fact that $\sqrt{a}-\sqrt{b}=\dfrac{a-b}{\sqrt{a}+\sqrt{b}}$ , we have
$$\begin{aligned}2\sqrt{x+1}-\frac{\sqrt{x}}{2}&=\frac12\left(4\sqrt{x+1}-\sqrt{x}\right)\\
&=\frac12\cdot\frac{16(x+1)-x}{4\sqrt{x+1}+\sqrt{x}}\\
&=\frac{15x+16}{2(4\sqrt{x+1}+\sqrt{x})}\\
&\geqslant\dfrac{15(x+1)}{10\sqrt{x+1}}\\
&=\dfrac32\sqrt{x+1}\\
&\geqslant\frac32\sqrt{2}\\
&=\sqrt{\frac92}>2 \end{aligned}$$
for $x\in\mathbb{N}$, so in fact the inequality is strict.
