Prove inequality $\left| \frac{x+yz}{x^2+y^2} \right| \leq 1$ for $x^2+y^2-z^2=1$

Question

Prove this:

If $x,y,z \in \mathbb R$ and $x^2+y^2-z^2=1$, then $$\left| \frac{x+yz}{x^2+y^2} \right| \leq 1 $$ holds.

Own ideas:

If we eliminate $z$ the inequality is equivalent to $$\left| \frac{x \pm y\sqrt{x^2+y^2-1}}{x^2+y^2} \right| \leq 1.$$

Answer 1

$(1+z^2)(x^2+y^2)\geq(x+yz)^2$ by Cauchy Inequality

Since $(1+z^2)=(x^2+y^2)$ the result follows.

Answer 2

WLOG, let $z=\tan C$ and $x=\sec C\cos B,y=\sec C\sin B$

$$\dfrac{x+yz}{x^2+y^2}=\cos^2C(\sec C\cos B+\sec C\sin B\tan C)$$

$$=\cos C\cos B+\sin B\sin C=\cos(B-C)$$