Inequality From Today’s Contest – Mattia Giuri's bizarre blog

Today a math contest I organized was held and I think this was the most difficult exercise overall.

There are m distinct positive even integers and n distinct positive odd integers that add up to 1987. Find the maximum of 3m + 4n.

So we have this sum of integers

$a_1 + a_2 + ... + a_m + b_1 + b_2 + ... + b_n = S = 1987$

We can write that

$S \geq (2 + 4 + ... + 2m) + (1 + 3 +... + 2n-1) = m^2 + m + n^2$

Now we can exploit the Cauchy-Schwartz inequality to make our desired coefficients appear.

$1987 + \frac{1}{4} \geq (m + \frac{1}{2})^2 + n^2$

$(3^2+4^2)(1987 + \frac{1}{4}) \geq (3^2+4^2)((m + \frac{1}{2})^2 + n^2) \geq (3m + \frac{3}{2} + 4n)^2$

Hence,

$3m + 4n \leq 5 \sqrt{1987 + \frac{1}{4}} - \frac{3}{2} < 222$

Hence the maximum is 221, which can be achieved with

$m=27, n=35$