<s>[INST] 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 $$\left| \frac{x \pm y\sqrt{x^2+y^2-1}}{x^2+y^2} \right| \leq 1.$$
 [/INST] $(1+z^2)(x^2+y^2)\geq(x+yz)^2$ by Cauchy Inequality
Since $(1+z^2)=(x^2+y^2)$ the result follows.
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