<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] 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 $x>0$
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