Show that (an + bn)1/n + (bn + cn)1/n + (cn + an)1/n < 1 + (21/n)/2, where n > 1 is an integer and a, b, c are the sides of a triangle with unit perimeter.
Solution
Thanks to Olena Bormashenko
We may take a ≥ b ≥ c. Since a + b + c = 1 and a < b+c, we have b ≤ a < 1/2. Hence (an + bn)1/n < 21/n/2 (*).
We have (b + c/2)n = bn + n/2 c bn-1 + other positive terms > bn + cn. Hence (bn + cn)1/n < b + c/2. Similarly, (cn + an)1/n < a + c/2. Adding we get (bn + cn)1/n + (cn + an)1/n < a+b+c = 1. Adding to (*) gives the required result.
© John Scholes
jscholes@kalva.demon.co.uk
24 Nov 2003
Last corrected/updated 24 Nov 03