For $a>1$, show that $\frac{1}{1+x}-\frac{1}{1+ax} \leq \frac{\sqrt{a}-1}{\sqrt{a}+1}$, $x \geq 1$ I'm self-learning the analysis I "by Herbert Amann" and got stuck in this problem. It's in Chapter IV Taylor's theorem.

For $a>1$ and $x\geq 1$ show that $$\frac{1}{1+x}-\frac{1}{1+ax} \leq \frac{\sqrt{a}-1}{\sqrt{a}+1}$$

This is what I've tried:
Let $f(t)=\frac{1}{1+tx}$ which is convex and,
$f(a) \geq f(1)+f'(1)(a-1)$
so $\frac{1}{1+x}-\frac{1}{1+ax} \leq \frac{x}{(1+x)^2}(a-1) \leq \frac{a-1}{4}$
 A: Isn't Taylor an overkill here? Given is equivalent to
$$\frac{x(a-1)}{(1+x)(1+ax)} \leq \frac{a-1}{(\sqrt{a}+1)^2}$$ or
$$x(\sqrt{a}+1)^2\leq {(1+x)(1+ax)}$$ or
$$xa+2x\sqrt{a}+x \leq 1+x+ax+ax^2$$
or $$2x\sqrt{a} \leq 1+ax^2$$
which is true since $$0\leq (1-x\sqrt{a})^2$$ and we didn' t even use $x>1$, but it must be $> -{1\over a}$.
A: Let
$$f : x \mapsto \frac{1}{1+x} - \frac{1}{1+ax}$$
$f$ is differentiable on $[1, +\infty)$ and
$$f'(x)= -\frac{1}{(1+x)^2}+\frac{a}{(1+ax)^2}$$
so $$f'(x)\geq 0 \Leftrightarrow \frac{a}{(1+ax)^2} \geq \frac{1}{(1+x)^2}$$
$$\Leftrightarrow \sqrt{a}(1+x) \geq 1+ax \Leftrightarrow \frac{1-\sqrt{a}}{\sqrt{a}-a} \geq x$$
So the maximum of $f$ is
$$f\left( \frac{1-\sqrt{a}}{\sqrt{a}-a}\right) = \frac{1}{1+\frac{1-\sqrt{a}}{\sqrt{a}-a}} - \frac{1}{1+a\frac{1-\sqrt{a}}{\sqrt{a}-a}} = \frac{\sqrt{a}-a}{1-a} - \frac{1-\sqrt{a}}{1-a}$$ $$= \frac{a+1-2\sqrt{a}}{a-1}= \frac{(\sqrt{a}-1)(\sqrt{a}-1)}{(\sqrt{a}+1)(\sqrt{a}-1)}=\frac{\sqrt{a}-1}{\sqrt{a}+1}$$
A: We need to prove that $$\frac{ax-x}{(1+ax)(1+x)}\leq\frac{\sqrt{a}-1}{\sqrt{a}+1}$$ or
$$\frac{x(\sqrt{a}+1)}{(1+ax)(x+1)}\leq\frac{1}{\sqrt{a}+1},$$ which is true by C-S:
$$\frac{x(\sqrt{a}+1)}{(1+ax)(x+1)}\leq\frac{x(\sqrt{a}+1)}{(\sqrt{x}+\sqrt{ax})^2}=\frac{1}{\sqrt{a}+1}.$$
