Given reals a1 ≤ a2 ≤ a3 ≤ a4 ≤ a5 ≤ a6 satisfying a1 + a2 + a3 + a4 + a5 + a6 = 10 and (a1 - 1)2 + (a2 - 1)2 + (a3 - 1)2 + (a4 - 1)2 + (a5 - 1)2 + (a6 - 1)2 = 6, what is the largest possible a6?
Solution
Answer: a6 = 3 1/3.
Adding twice the first equation to the second gives a12 + a22 + ... + a62 = 20. Hence (a1 - 4/3)2 + (a2 - 4/3)2 + (a3 - 4/3)2 + (a4 - 4/3)2 + (a5 - 4/3)2 + (a6 - 4/3)2 = 20 - 10·8/3 + 6·16/9 = 4. So a6 - 4/3 ≤ 2, or a6 ≤ 3 1/3. It is easy to check that a1 = a2 = a3 = a4 = a5 = 4/3, a6 = 3 1/3 is indeed a solution.
Comment. Note that this is similar to 85/2
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002