Finding the maximum value without using derivatives 
Find the maximum value of $$f(x)=2\sqrt{x}-\sqrt{x+1}-\sqrt{x-1}$$ without using derivatives.

The domain of $f(x)$ is $x \in [1,\infty)$. Then, using derivatives, I can prove that the function decreases for all $x$ from $D(f)$ and the maximum value is $f(1)= 2 - \sqrt{2}$. However, this uses derivatives.
 A: Some other manipulation perhaps?
$$\begin{align}
f(x)&=2\sqrt{x}-\sqrt{x+1}-\sqrt{x-1} \\
&=\sqrt{x}\left[2-\frac{\sqrt{x+1}}{\sqrt{x}}-\frac{\sqrt{x-1}}{\sqrt{x}}\right] \\
&=\sqrt{x}\left[2-\left(\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x}}\right)\right] 
\end{align}$$
With $\left(\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x}}\right) \to 2$ shows that $\left[2-\left(\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x}}\right)\right]$ is very close to $0$ for larger $x \ \  $ , letting you look at smaller values that take away the diminishing effect of $\left(\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x}}\right) $
A: Note that
\begin{align}
f(x) & = (\sqrt{x}-\sqrt{x-1})-(\sqrt{x+1}-\sqrt{x})\\
& = \frac{1}{\sqrt{x}+\sqrt{x-1}} - \frac{1}{\sqrt{x+1}+\sqrt{x}}\\
& = \frac{\sqrt{x+1}-\sqrt{x-1}}{(\sqrt{x}+\sqrt{x-1})(\sqrt{x+1}+\sqrt{x})} \\
& = \frac{2}{(\sqrt{x}+\sqrt{x-1})(\sqrt{x+1}+\sqrt{x})(\sqrt{x+1}+\sqrt{x-1})} 
\end{align}
which is a decreasing function of $x$.
A: $$f(x)=\frac{1}{\sqrt{x}+\sqrt{x-1}}-\frac{1}{\sqrt{x+1}+\sqrt{x}}=$$
$$=\frac{2}{(\sqrt{x}+\sqrt{x-1})(\sqrt{x}+\sqrt{x+1})(\sqrt{x+1}+\sqrt{x-1})}\leq$$
$$=\frac{2}{(1+\sqrt2)\sqrt2}=2-\sqrt2.$$
The equality occurs for $x=1$, which says that we got a maximal value.
