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 &quot;by Herbert Amann&quot; and got stuck in this problem. It's in Chapter IV Taylor's theorem.

For $a&gt;1$ 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 $\frac{1}{1+x}-\frac{1}{1+ax} \leq \frac{x}{(1+x)^2}(a-1) \leq \frac{a-1}{4}$
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 $x&gt;1$, but it must be $&gt; -{1\over a}$.
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:
$$\frac{x(\sqrt{a}+1)}{(1+ax)(x+1)}\leq\frac{x(\sqrt{a}+1)}{(\sqrt{x}+\sqrt{ax})^2}=\frac{1}{\sqrt{a}+1}.$$