Given $r_{i}>0;\: i=1\ldots K$ when $\frac{\sum_{i=1}^{K}l_{i}}{max(\frac{l_{1}}{r_{1}},\ldots,\frac{l_{K}}{r_{k}})}$ is maximized?

Question

Given $r_{i}>0;\: i=1\ldots K$ , We want to determine $l_{i}>0;\: i=1,\ldots,K$, such that function $f(l_{1},\ldots,l_{K})=\frac{\sum_{i=1}^{K}l_{i}}{\max\left(\frac{l_{1}}{r_{1}},\ldots,\frac{l_{K}}{r_{K}}\right)}$ is maximized. How can we show $f(l_{1},\ldots,l_{K})$ is maximized when the relation $\frac{l_{1}}{r_{1}}=\frac{l_{2}}{r_{2}}=\cdots=\frac{l_{K}}{r_{K}}$ is hold?

Answer 1

A possible simplification of the problem.

Let $l_i=m_i r_i$ then we have:

$$g(m_{1},\ldots,m_{K})=f(m_{1}r_{1},\ldots,m_{K}r_{K})=\frac{\sum_{i=1}^{K}m_{i}r_{i}}{\max\left(m_{1},\ldots,m_{K}\right)}$$.

The problem then becomes:

Given $r_{i}>0;\: i=1\ldots K$ , We want to determine $m_{i}>0;\: i=1,\ldots,K$, such that function $g(m_{1},\ldots,m_{K})=\frac{\sum_{i=1}^{K}m_{i}r_i}{\max\left(m_{1},\ldots,m_{K}\right)}$ is maximized. How can we show $g(m_{1},\ldots,m_{K})$ is maximized when the relation $m_{1}=m_{2}=\cdots=m_{K}$ is hold?