Prove $ \forall x >0, \quad \sqrt{x +2} - \sqrt{x +1} \neq \sqrt{x +1}-\sqrt{x}$ 
I would like to prove
  $$ \forall x >0,  \quad \sqrt{x +2} - \sqrt{x +1} \neq \sqrt{x +1}-\sqrt{x}$$
  
  
*
  
*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.


Assume the statement is true
\begin{align}
\sqrt{x + 2} - \sqrt{x + 1} &= \sqrt{x + 1} - \sqrt{x}\\
\sqrt{x + 2} + \sqrt{x} &= 2\sqrt{x + 1}\\
2x + 2 + (2\sqrt{x^2 + 2x}) &= 4x + 4\\
\sqrt{x^2 + 2x} &= x + 1\\
x^2 + 2x &= x^2 + 2x + 1\\
0 &= 1\\
\end{align}
Which is a contradiction. Now I am not sure if I'm allowed to manipulate the equation like this so the proof might be invalid. The reason I though I could was I had seen simillar logic in proving the irrationality of $\sqrt{2}$. 
