ABCD is a tetrahedron with centroid G. The line AG meets the circumsphere again at A'. The points B', C' and D' are defined similarly. Show that GA.GB.GC.GD ≤ GA'.GB'.GC'.GD' and 1/GA + 1/GB + 1/GC + 1/GC ≥ 1/GA' + 1/GB' + 1/GC' + 1/GD'.
Solution
Let O be the circumcenter of the tetrahedron and R the circumradius. Take vectors origin O. Denote OP by P etc. Then GA2 = (G - A)2 = G2 + A2 - 2G.A = OG2 + R2 - 2G.A. Adding the similar equations for B, C, D we get GA2 + GB2 + GC2 + GD2 = 4 OG2 + 4 R2 - 2G.(4 G) = 4(R2 - OG2). So by the arithmetic/geometric mean inequality we have (GA2GB2GC2GD2) ≤ (R2 - OG2)4.
But we also have GA.GA' = R2 - OG2 etc. So (R2 - OG2)4 = (GA GB GC GD)(GA' GB' GC' GD'). Hence GA.GB.GC.GD ≤ GA'.GB'.GC'.GD', which is the first inequality.
Cauchy-Schwartz gives (GA.1 + GB.1 + GC.1 + GD.1)2 ≤ (GA2 + GB2 + GC2 + GD2)(12 + 12 + 12 + 12) and 16 = (GA1/2.(1/GA)1/2 + GB1/2.(1/GB)1/2 + GC1/2.(1/GC)1/2 + GD1/2.(1/GD)1/2)2 ≤ (GA + GB + GC + GD)(1/GA + 1/GB + 1/GC + 1/GD). All sums are positive, so multiplying the two inequalities we get 1/4 (GA2 + GB2 + GC2 + GD2)(1/GA + 1/GB + 1/GC + 1/GD) ≥ (GA + GB + GC + GD). Hence (1/GA + 1/GB + 1/GC + 1/GD) ≥ (GA + GB + GC + GD)/(R2 - OG2) = 1/GA' + 1/GB' + 1/GC' + 1/GD', which is the second inequality.
© John Scholes
jscholes@kalva.demon.co.uk
9 Oct 2002