Find the maximum in integers for n,m: n/(n+m) + (Nn)/(Nn + Mm)

Question

Mathematics

I messed up something. I feel that the solution is simple, but I can not immediately solve it. Help, mathematician :).

Find the maximum f(n,m) for non-negative integers n,m: f(n,m) = n/(n+m) + (Nn)/(Nn + Mm)
where N>=n, M>=m, M,N>0

Thank you.

Answer 1

I made a change of variables (x = n, y = n + m, X = N, Y = N + M) and differentiated. As a result, the maximum is reached at n + m = (N + M) / 2. In the case of an even N + M, the maximum of the function is 2 * N / (N + M)

Answer 2

Try to write everything honestly and calculate, the upper part of the fraction
is 2 n^2 + (2 N + M - 2 m) n + mN - > max (you can factorize and find the maximum with respect to n)
The lower part is (n + m) ( (N + M) - (n + m)), is naturally minimized at n + m = (N + M), although division by 0 occurs here. That is, if the upper part is the maximum and the lower part is the minimum and these values ​​are consistent, then there is an answer :) Naturally, the maximum for obvious reasons is less than 2

Answer 3

Well, if the maximum that we can squeeze out of n / (n + m) is 1, with m = 0. Going further, (Nn) / (Nn + Mm) here the best case is the one where the difference M - m is minimal, then is 1. Therefore, for M = 1, m = 0, n = 1, N = INF, the limit of (Nn)/(Nn + Mm) is equal to one, which means that the maximum of the function tends to 2.