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.


By the mean value theorem - MVT, for all $x >0$, it exists $\zeta_x \in (x, x+1)$ such that $$\frac{1}{2\sqrt{\zeta_x}}((x+1)-x)=\frac{1}{2\sqrt{\zeta_x}}=\sqrt{x+1}-\sqrt{x}$$ As $y \to \frac{1}{2\sqrt{y}}$ is stricly decreasing, you have $$\frac{1}{2\sqrt{\zeta_{x+1}}}=\sqrt{x+2} -\sqrt{x+1} < \sqrt {x+1}-\sqrt{x}=\frac{1}{2\sqrt{\zeta_x}}.$$
