a_{ij} are reals in [0, 1]. Show that ( ∑_{i=1}^{n} ∑_{j=1}^{mi} a_{ij}/i )^{2} ≤ 2m ∑_{i=1}^{n} ∑_{j=1}^{mi} a_{ij}.

**Solution**

The question looks incredibly complicated, but is actually easy. We just use induction and the fact that each a_{ij} ≤ 1.

Put b_{i} = ∑_{1}^{mi} a_{ij}. Notice that since each a_{ij} ≤ 1, we have b_{i} ≤ mi. We use induction on n. For n = 1, the required result is: b_{1}^{2} ≤ 2mb_{1}. But b_{1} ≤ m and b_{1} ≥ 0, so this is certainly true.

Now assume the result is true for n.

For n + 1, the lhs is (b_{1}/1 + b_{2}/2 + ... + b_{n}/n)^{2} + 2(b_{1}/1 + ... + b_{n}/n) b_{n+1}/(n+1) + b_{n+1}^{2}/(n+1)^{2}, and the rhs is 2m(b_{1} + ... + b_{n}) + 2mb_{n+1}. Given the result for n, it is sufficient to show that: 2(b_{1}/1 + ... + b_{n}/n) b_{n+1}/(n+1) + b_{n+1}^{2}/(n+1)^{2} ≤ 2mb_{n+1}. Divding by b_{n+1} and using b_{i}/i le; m, we need: 2nm/(n+1) + m/(n+1) ≤ 2m, which is clearly true.

© John Scholes

jscholes@kalva.demon.co.uk

30 Nov 1999