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.


I will show that
if
$\sqrt{x+a}-\sqrt{x+b}
=\sqrt{x+b}-\sqrt{x+c}
$
for two different values of $x$,
then
$a=c$
and
$b = a+c
=2a
$.
If the equality
holds for one value of $x$,
then
$8x(a+c-2b)
=(a-4b+c)^2-4ac
$.
Since,
in the original problem,
$a=2, b=1, c=0$,
this can not hold for
two values of $x$.
If it holds for one $x$,
since
$a+c-2b = 0$
and
$(a-4b+c)^2-4ac
=2^2-0
=4
$,
this does not hold for
any $x$.
then
Suppose
$\sqrt{x+a}-\sqrt{x+b}
=\sqrt{x+b}-\sqrt{x+c}
$,
or
$\sqrt{x+a}+\sqrt{x+c}
=2\sqrt{x+b}
$.
Squaring,
$x+a+x+c+2\sqrt{(x+a)(x+c)}
=4(x+b)
$
or
$2x-a+4b-c
=2\sqrt{(x+a)(x+c)}
$.
Squaring again,
$4x^2-4x(a-4b+c)+(a-4b+c)^2
=4(x+a)(x+c)
=4(x^2+x(a+c)+ac)
=4x^2+4x(a+c)+4ac
$
or
$4x(a+c+a-4b+c)
=(a-4b+c)^2-4ac
$
or
$4x(2a+2c-4b)
=(a-4b+c)^2-4ac
$
or
$8x(a+c-2b)
=(a-4b+c)^2-4ac
$.
If this is true for
more than one $x$,
then
$a+c-2b = 0$
and
$(a-4b+c)^2-4ac
= 0
$.
From the first equation,
$2b = a+c$.
Substituting this
in the second equation,
$4ac
=(a-4b+c)^2
=(a+c-2(a+c))^2
=(a+c)^2
=a^2+2ac+c^2
$
or
$0
=a^2-2ac+c^2
=(a-c)^2
$,
so that
$a=c$.
Therefore, 
if
$\sqrt{x+a}-\sqrt{x+b}
=\sqrt{x+b}-\sqrt{x+c}
$
for two different values of $x$,
then
$a=c$
and
$b = a+c
=2a
$.
