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.

You can set $f(x)=2\sqrt{x+1}-2-\frac 12\sqrt{x}$
The derivative $f'(x)=\frac 1{\sqrt{x+1}}-\frac 1{4\sqrt{x}}$
It is zero when $(4\sqrt{x})^2=(\sqrt{x+1})^2\iff 16x=x+1\iff x=\frac 1{15}$
The derivative is $-$ then $+$ so $f$ is $\searrow$ then $\nearrow$ with a minimum in $\frac 1{15}$.
We could calculate $f(\frac 1{15})$ but since we are only interested in naturals let just calculate
$f(1)=2\sqrt{2}-2-\frac 12\sqrt{2}>0$ as you have noticed yourself.
And since it is increasing, it's true for all $x\ge 1$ not only integers.

For sake of completion, $f(\frac 1{15})=\frac 12\sqrt{15}-2<\frac 12\sqrt{16}-2=0$
So there exists a zero of $f$ in the interval $[\frac 1{15},1]$ in addition to $x=0$.
We can calculate it by squaring both sides:
$(2\sqrt{x+1})^2=(2+\frac 12\sqrt{x})^2\iff 4x+4=4+\frac 14x+2\sqrt{x}\iff\frac {15}4x=2\sqrt{x}\iff \\\sqrt{x}=\frac{8}{15}\text{ or }x=0$
So actually the inequality is true for all $x\ge \frac{64}{225}$
Side note: this direct calculation of the zero, shows there are only two zeroes, $x=0$ and $x=\frac{64}{225}$, so you can deduce directly (without derivative) that since $f(1)>0$ and $f$ is continuous then $f$ cannot change sign on $[1,+\infty)$
