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.

 A: Your proof is correct, but I feel that this could be proved by contradiction.
Assume for contradiction $\exists x>0$ such that the equation $\sqrt{x+2}-\sqrt{x+1}=\sqrt{x+1}-\sqrt{x}$ is true. Then,
\begin{align}
\sqrt{x+2}-\sqrt{x+1}&=\sqrt{x+1}-\sqrt{x}\\
\frac{x+2-x-1}{\sqrt{x+2}+\sqrt{x+1}}&=\frac{x+1-x}{\sqrt{x +1}+\sqrt{x}}\\
\frac{1}{\sqrt{x+2}+\sqrt{x+1}}&=\frac{1}{\sqrt{x +1}+\sqrt{x}}\\
\sqrt{x +1}+\sqrt{x} &=\sqrt{x+2}+\sqrt{x+1}\\
\sqrt{x} &=\sqrt{x+2}\\
x&=x+2
\end{align}
This is not true and we have reached a contradiction. Thus the equation does not hold.
A: 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}
A: Hint #1:
Assume that $\sqrt{x + 2} - \sqrt{x + 1} = \sqrt{x + 1} - \sqrt{x}$ for some $x > 0$.
Hint #2:
Derive a contradiction.
Hint #3:
This proof (by contradiction) results to some changes in the notation you used in your proof.
A: 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 $x\ge 0.$
A: 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 $\sqrt{2}$. 
A: 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 $$\frac{1}{2\sqrt{\zeta_{x+1}}}=\sqrt{x+2} -\sqrt{x+1} < \sqrt {x+1}-\sqrt{x}=\frac{1}{2\sqrt{\zeta_x}}.$$
A: 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
$.
