Question: There is a set of numbers whose sum is equal to the sum of the elements squared. What's bigger: the sum of the cubes or the sum of the fourth powers?
This is a question taken from a set of past Oxbridge interview questions.
The progress I've made is:
- A trivial solution is just a set comprised of 1s in which case the sums would always be equal
- This set of numbers should include a mix of numbers with absolute values less than 1 and also greater than 1
- One example of such a set is $(1.5,0.5,0.5,0.5)$ which gives 3 for both the sum and sum of squares. In this example, the sum of the fourth powers is greater.
- Intuitively I reason that the sum of fourth powers should be greater because the elements with absolute value less than 1 can only really change their absolute value from 1 to 0 whereas the elements with absolute value >1 increase without bound hence the fourth powers should be greater.
My issue with this is that my solution is too loose, if anyone could provide better reasoning I would be grateful.
If the numbers are assumed to be positive, use the fact that $\sum_i x_i^\alpha$ is a convex function of $\alpha$.
If they are allowed to be both positive and negative, then either could be bigger. Compare the cases $\{2/5, 6/5\}$ and $\{-1/5, 2/5\}$.