Solving $|y-\sqrt{2}| \leq (\sqrt{2} - 1) |x-\sqrt{2}|$

Question

I am an undergraduate student and currently I am working on inequalities and the properties of absolute values. I am trying to solve the following problem, but I do not know how to approach it:

Being $x >= 0$. Suppose that $y = \frac{x+2}{x+1}$. Calculate $\frac{y-\sqrt{2}}{x-\sqrt{2}}$. Deduce the following inequality:

$$|y-\sqrt{2}| \leq (\sqrt{2} - 1) |x-\sqrt{2}|$$

Answer 1

Show that

$$\frac{y-\sqrt{2}}{x-\sqrt{2}} = - \frac{ \sqrt{2} - 1 } { x + 1 }. $$

(This is the numerical answer to the intermediate step. The proof is just by expanding the algebra and factoring.)

Hence, conclude that

$$ \left| \frac{y-\sqrt{2}}{x-\sqrt{2}} \right| = \frac{ \sqrt{2} - 1 } { x + 1 } \leq \sqrt{2} - 1 . $$