<s>[INST] 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.
 [/INST] $$\sqrt{x} - \sqrt{x-1}>\sqrt{x+1} - \sqrt{x}\quad| +\sqrt{x}$$
$$2\sqrt{x} > \sqrt{x-1}+\sqrt{x+1} \quad|^2$$
$$4x>2x+ 2\sqrt{(x-1)(x+1)} \quad| -2x $$
$$2x>2\sqrt{(x-1)(x+1)} \quad| /2$$
$$x=\sqrt{x^2-1} \quad|^2$$
$$x^2>x^2-1 \quad|-x^2$$
$$0>-1$$
The transfoemations are valid in both directions, top down and bottom up
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