<s>[INST] 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!
 [/INST] \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 $x+2 > x$.
</s>