<s>[INST] Prove $ \forall x >0, \quad \sqrt{x +2} - \sqrt{x +1} \neq \sqrt{x +1}-\sqrt{x}$ 
I would like to prove
  $$ \forall x >0,  \quad \sqrt{x +2} - \sqrt{x +1} \neq \sqrt{x +1}-\sqrt{x}$$
  
  
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*I'm interested in more ways of proving it
  

My thoughts:
\begin{align}
\sqrt{x+2}-\sqrt{x+1}\neq \sqrt{x+1}-\sqrt{x}\\
\frac{x+2-x-1}{\sqrt{x+2}+\sqrt{x+1}}&\neq \frac{x+1-x}{\sqrt{x +1}+\sqrt{x}}\\
\frac{1}{\sqrt{x+2}+\sqrt{x+1}}&\neq \frac{1}{\sqrt{x +1}+\sqrt{x}}\\
\sqrt{x +1}+\sqrt{x} &\neq \sqrt{x+2}+\sqrt{x+1}\\
\sqrt{x} &\neq \sqrt{x+2}\\
\end{align}


*

*Is my proof correct? 

*I'm interested in more ways of proving it.

 [/INST] By the MVT:
$$\sqrt {x+2} - \sqrt {x+1} = \frac{1}{2\sqrt {c_x}}\cdot 1, \ \ \ \ \sqrt {x+1} - \sqrt {x} = \frac{1}{2\sqrt {d_x}}\cdot 1.$$
Here $c_x\in (x+1,x+2), d_x\in (x,x+1).$ Because $1/\sqrt x$ strictly decreases, the left term minus the right term is negative.

Concavity: Slopes of successive chords on a strictly concave graph are strictly decreasing, and $\sqrt x$ is strictly concave. Therefore $\sqrt{x +2} - \sqrt{x +1} < \sqrt{x +1}-\sqrt{x}$  for all $x\ge 0.$
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