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.


Your proof is correct, because each inequality you write is equivalent to the previous one (it should be noted, probably).
Changing all $\ne$ into $=$ would make it a proof by contradiction, that's however unnecessary.
In a different way, you could just swap terms and square, again changing inequalities into equivalent ones:
\begin{gather}
\sqrt{x+2}+\sqrt{x}\ne 2\sqrt{x+1}\\[10px]
x+2+x+2\sqrt{x(x+2)}\ne 4x+4\\[10px]
\sqrt{x(x+2)}\ne x+1\\[10px]
x^2+2x\ne x^2+2x+1\\[10px]
0\ne1
\end{gather}
