China Southeastern Mathematical Olympiad

Note that
$$\begin{aligned} 1 &= x^2 + y^2 + z^2 \\ &= x^2 + \frac{\lambda^2}{1+\lambda^2}y^2 + \frac{1}{1+\lambda^2}y^2 + z^2 \\ &\ge \frac{2}{\sqrt{1+\lambda^2}}(|x| + |y| + |z|) \\ &\ge \frac{2}{\sqrt{1+\lambda^2}}(|\lambda xy + yz|), \end{aligned}$$
and the two equalities hold simultaneously when
$$y = \frac{\sqrt{2}}{2}, \quad x = \frac{\sqrt{2}\lambda}{2\sqrt{\lambda^2+1}}, \quad z = \frac{\sqrt{2}}{2\sqrt{\lambda^2+1}}.$$
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 $\lambda = 2$.