How to prove the following inequality holds:

Question

Let $$y \geq -D$$$$ 0 \leq x < \frac{1}{D}$$ where $D > 0.$ How do I show that $$\frac{y}{1 + xy} \geq \frac{1}{x-\frac{1}{D}}$$

Answer 1

$f(y) = \dfrac{y}{1+xy} \Rightarrow f'(y) = \dfrac{1}{(1+xy)^2} > 0 \Rightarrow f(y) \geq f(-D) = \dfrac{1}{x-\dfrac{1}{D}}$