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 $\frac{1}{1+x}-\frac{1}{1+ax} \leq \frac{x}{(1+x)^2}(a-1) \leq \frac{a-1}{4}$

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:
$$\frac{x(\sqrt{a}+1)}{(1+ax)(x+1)}\leq\frac{x(\sqrt{a}+1)}{(\sqrt{x}+\sqrt{ax})^2}=\frac{1}{\sqrt{a}+1}.$$
