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.

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 $x$.
• @Make42 1. Consider $(a+b)(a-b)=a^2-b^2$ and $\sqrt x^2-\sqrt{x-1}^2=x-(x-1)=1$. 2. All terms in the denominator are non-negative and increasing hence their sums are positive and increasing, too, and so is the product of the sums. Then its reciprocal is decreasing. – CiaPan Feb 6 '18 at 11:27
$$f(x)=\frac{1}{\sqrt{x}+\sqrt{x-1}}-\frac{1}{\sqrt{x+1}+\sqrt{x}}=$$ $$=\frac{2}{(\sqrt{x}+\sqrt{x-1})(\sqrt{x}+\sqrt{x+1})(\sqrt{x+1}+\sqrt{x-1})}\leq$$ $$=\frac{2}{(1+\sqrt2)\sqrt2}=2-\sqrt2.$$ The equality occurs for $x=1$, which says that we got a maximal value.
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 $\left(\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x}}\right)$