XXI Geometrical Olympiad in Honor of I.F. Sharygin, correspondence round, problem 24
Tags: inequality 3D tetrahedron spheres insphere
We will utilize the known fact that \(\angle AC_1B = \angle AD_1B = \angle CA_1D = \angle CB_1D\) (and other similar equalities). Denote \(AB_1=AC_1=AD_1=a\) and similarly define \(b, c, d\). Also denote \(\angle AD_1B=\gamma, \angle BD_1C = \alpha, \angle CD_1A = \beta\) (clearly \(\alpha, \beta, \gamma < \pi\)). Then \[S_{AB}S_{CD} = \frac{1}{2}ab\sin\gamma \cdot \frac{1}{2}cd\sin\gamma = \frac{1}{4}abcd\sin^2\gamma.\] Thus, triangle inequality \(\sqrt{S_{AB} S_{CD}} + \sqrt{S_{AC} S_{BD}} > \sqrt{S_{AD} S_{BC}}\) is equivalent to the inequality \(\sin\gamma + \sin\beta>\sin\alpha\). To prove this, mark points \(A_2, B_2, C_2\) on rays \(D_1A, D_1B\) and \(D_1C\), respectively, at distance \(1\) from \(D_1\). Then the doubled areas of \(A_2B_2D_1\), \(A_2C_2D_1\) and \(B_2C_2D_1\) are equal to \(\sin\gamma, \sin\beta\) and \(\sin\alpha\), respectively. \(D_1\) is the circumcenter of \(A_2B_2C_2\) and lies inside it, consequently lies inside its medial triangle. Hence the area of \(B_2C_2D_1\) is less than half the area of \(A_2B_2C_2\), i.e. \(\sin\gamma+\sin\beta>\sin\alpha\), as desired.