China Southeastern Mathematical Olympiad

Note that $$\begin{array}{rl}1& =x^2+y^2+z^2\\ & =x^2+\frac{{\lambda }^2}{1+{\lambda }^2}{y}^2+\frac{1}{1+{\lambda }^2}{y}^2+z^2\\ & \ge \frac{2}{\sqrt{1+{\lambda }^2}}(|x|+|y|+|z|)\\ & \ge \frac{2}{\sqrt{1+{\lambda }^2}}(|\lambda xy+yz|),\end{array}$$ and the two equalities hold simultaneously when $$y=\frac{\sqrt{2}}{2},x=\frac{\sqrt{2}\lambda }{2\sqrt{{\lambda }^2+1}},z=\frac{\sqrt{2}}{2\sqrt{{\lambda }^2+1}}.$$ Thus, $\frac{\sqrt{1+{\lambda }^2}}{2}$ is the maximum value of $|\lambda xy+yz|$. Let $$\frac{\sqrt{1+{\lambda }^2}}{2}=\frac{\sqrt{5}}{2}.$$ We obtain that $\lambda =2$.