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.

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.
