Question 3
(BUL 6; IMO 5) In the tetrahedron $SABC$ the angle $BSC$ is a right angle, and the projection of the vertex $S$ to the plane $ABC$ is the intersection of the altitudes of the triangle $ABC$. Let $z$ be the radius of the inscribed circle of the triangle $ABC$. Prove that
$$SA^2+SB^2+SC^2\ge18z^2.$$
$$SA^2+SB^2+SC^2\ge18z^2.$$