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

WLOG, let $z=\tan C$  and $x=\sec C\cos B,y=\sec C\sin B$
$$\dfrac{x+yz}{x^2+y^2}=\cos^2C(\sec C\cos B+\sec C\sin B\tan C)$$
$$=\cos C\cos B+\sin B\sin C=\cos(B-C)$$
