Let $SABC$ be the pyramid where$ m(\angle ASB) = m(\angle BSC) = m(\angle CSA) = 90^o$. Show that, whatever the point $M \in AS$ is and whatever the point $N \in BC$ is, holds the relation $$\frac{1}{MN^2} \le \frac{1}{SB^2} + \frac{1}{SC^2}.$$
Source: 2000 Romania NMO VIII p3
Tags: geometry, 3D geometry, pyramid
Let $SABC$ be the pyramid where$ m(\angle ASB) = m(\angle BSC) = m(\angle CSA) = 90^o$. Show that, whatever the point $M \in AS$ is and whatever the point $N \in BC$ is, holds the relation $$\frac{1}{MN^2} \le \frac{1}{SB^2} + \frac{1}{SC^2}.$$