How to prove $\sqrt{x} - \sqrt{x-1}>\sqrt{x+1} - \sqrt{x}$ for $x\geq 1$? Intuitively when $x$ gets bigger, $\sqrt{x+1}$ will get closer to $\sqrt{x}$, so their difference will get smaller.
However, I just cannot get a proper proof.
 A: Note, for $x\ge 1$
$$x>\sqrt{x^2-1}$$
$$4x>2x+2\sqrt{x^2-1}=(\sqrt{x+1}+\sqrt{x-1})^2$$
$$2\sqrt x>\sqrt{x+1}+\sqrt{x-1} $$
$$\sqrt x -\sqrt{x-1}> \sqrt{x+1}-\sqrt x$$
A: Alternative approach that avoids Calculus:
square both sides.
LHS squared is $(2x-1) - 2\sqrt{x(x-1)}$ 
RHS squared is $(2x+1) - 2\sqrt{x(x+1)}$
So, taking LHS - RHS, the question reduces to whether 
$-2 + 2\sqrt{x}[\sqrt{x+1} - \sqrt{x-1}] > 0.$
This is equivalent to asking whether 
$\sqrt{x}[\sqrt{x+1} - \sqrt{x-1}] > 1.$
Edit
Stealing a caveat from Quanto's answer, note that when $x\geq 1$, the LHS above is clearly positive.
Again squaring both sides, the problem reduces to asking whether 
$x[2x - 2\sqrt{x^2 -1}] > 1.$
This is answered by applying the scalar of $[2x + 2\sqrt{x^2 -1}],$ which is a positive scalar,
to both sides, so that the problem is transformed to asking whether
$$x[2x - 2\sqrt{x^2 -1}][2x + 2\sqrt{x^2 -1}] > [2x + 2\sqrt{x^2 -1}].\tag1$$
Since the LHS of equation (1) above simplifies to $4x$,
and since it is clear that $2\sqrt{x^2 - 1} < 2x$ 
the problem is resolved.
A: \begin{align}
   \sqrt{x} - \sqrt{x-1} 
   &= \dfrac{(\sqrt{x} - \sqrt{x-1})(\sqrt{x} + \sqrt{x-1})}{\sqrt{x} + \sqrt{x-1})} \\
   &= \dfrac{x-(x-1)}{\sqrt{x} + \sqrt{x-1}} \\
   &= \dfrac{1}{\sqrt{x} + \sqrt{x-1}}
\end{align}
\begin{align}
   \sqrt{x+1} - \sqrt{x} 
   &= \dfrac{(\sqrt{x+1} - \sqrt{x})(\sqrt{x+1} + \sqrt{x})}{\sqrt{x+1} + \sqrt{x})} \\
   &= \dfrac{(x+1)-x}{\sqrt{x+1} + \sqrt{x}} \\
   &= \dfrac{1}{\sqrt{x+1} + \sqrt{x}}
\end{align}
So
\begin{align}
   \sqrt{x+1} &> \sqrt{x-1} \\
   \sqrt{x+1} + \sqrt x &> \sqrt x + \sqrt{x-1} \\
   \left(\dfrac{1}{\sqrt{x+1} + \sqrt x}\right) 
      &< \left(\dfrac{1}{\sqrt x + \sqrt{x-1}}\right) \\
   \sqrt{x+1} - \sqrt x &< \sqrt x - \sqrt{x-1} \\
\end{align}
A: $$\sqrt{x} - \sqrt{x-1}>\sqrt{x+1} - \sqrt{x}\quad| +\sqrt{x}$$
$$2\sqrt{x} > \sqrt{x-1}+\sqrt{x+1} \quad|^2$$
$$4x>2x+ 2\sqrt{(x-1)(x+1)} \quad| -2x $$
$$2x>2\sqrt{(x-1)(x+1)} \quad| /2$$
$$x=\sqrt{x^2-1} \quad|^2$$
$$x^2>x^2-1 \quad|-x^2$$
$$0>-1$$
The transfoemations are valid in both directions, top down and bottom up
A: This is not a proof.
When $x$ is large
$$\Big[\sqrt{x} - \sqrt{x-1}\Big]-\Big[\sqrt{x+1} - \sqrt{x}\Big]=\frac 1{4 x^{\frac 32}}\sum_{n=0}^\infty \frac {a_n}{x^{2n}}$$  all coefficients being positive make the decreasing sequence
$$\left\{1,\frac{5}{16},\frac{21}{128},\frac{429}{4096},\frac{2431}{32768},\frac{29393
   }{524288},\frac{185725}{4194304},\cdots\right\}$$
