Is there a neat way to find the sign of a real function with radicals like this? This simple function:
$$f(x) = \sqrt{x+2} - 2\sqrt{x+1} + \sqrt{x}$$
is negative for $x>0$ (just checked out with a graphic calculator). 
But how to "prove" this algebraically?
The function comes out from the problem of finding which of these numbers $\sqrt{12} - \sqrt{11}$ and $\sqrt{11} - \sqrt{10}$ is bigger. 
[BTW, given that $f(x)$ is negative, it should be that $\sqrt{11} - \sqrt{10}$ is bigger than $\sqrt{12} - \sqrt{11}$ !]
Thanks!
 A: \begin{align}
f(x) &= \sqrt{x+2}-2\sqrt{x+1}+\sqrt{x}\\
&= \sqrt{x+2}-\sqrt{x+1}-(\sqrt{x+1}-\sqrt{x})\\
&=\frac{1}{\sqrt{x+2}+\sqrt{x+1}}-\frac1{\sqrt{x+1}+\sqrt{x}}\\
&<0
\end{align}
since $x+2 > x$.
A: I will prove $f(x)<0$ 
So i need to prove $\sqrt{x+2}-2\sqrt{x+1}+\sqrt{x}<0\left(x>0\right)$
Or $\sqrt{x+2}+\sqrt{x}<2\sqrt{x+1}$
By square both sides we have: $\sqrt{x^2+2x}<x+1$
Or $x^2+2x<x^2+2x+1\Leftrightarrow0<1$ 
So $f(x)<0$ it means  $f(x)$ is negative for $x>0$
A: With some analysis
The square root function is strictly concave, hence
$$\sqrt{\tfrac12 x+\tfrac12(x+2)}=\sqrt{x+1}>\tfrac12\sqrt x+\tfrac12\sqrt{x+2}\iff \sqrt x+\sqrt{x+2}<2\sqrt{x+1}.$$
A: Here is a calculus-based answer, that I find more intuitive. We want to prove that
$$
\sqrt{x+1} - \sqrt{x} > \sqrt{x+2} - \sqrt{x + 1}.
$$
Fix $x > 0$. Then the fundamental theorem of calculus tells us that what we want to prove is equivalent to
$$
\int_{x}^{x+1} \frac1{2\sqrt{y}} \mathrm dy > \int_{x+1}^{x+2} \frac1{2\sqrt{y}} \mathrm dy.
$$
But because the function $\sqrt{y}$ is strictly increasing, the function $\frac1{2\sqrt{y}}$ is strictly decreasing. In particular, integrating over an interval of length one gives lower numbers as the lower bound of said interval increases.
A: $(x+2)^{1/2}-(x+1)^{1/2}  - ((x+1)^{1/2} -x^{1/2})$.
MVT:
$(x+k+1)^{1/2}-(x+k)^{1/2}=$
$(1/2)(\frac{1}{a})^{1/2}×1,$ where
$a \in (x+k, x+k+1)$.
Compare the terms for $k=0,1$
Used: 
$\dfrac {f(x)-f(x_o)}{x-x_o}=f'(a)$, where
$a \in (\min(x_0,x),\max(x_0,x))$.
