<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] \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}
</s>