Let $\mu_i\in [0,1]$ for $i=1,...,N$ and $\sum_{i=1}^N \mu_i=1$. Let $t_i\in [0,1]$ for $i=1,...,N$. Consider the function $$ f(\mu_i,t)=\frac{\mu_i t_i }{\sum_{j=1}^n \mu_j t_j} $$ Take two values of $\mu$, e.g., $\bar{\mu}=(\bar{\mu}_1,\bar{\mu_2},...,\bar{\mu}_N)$ and $\tilde{\mu}\equiv (\tilde{\mu}_1,\tilde{\mu_2},...,\tilde{\mu}_N)$.
Is it true that $$ \bar{\mu_1}\leq \tilde{\mu_1} \Rightarrow f(\bar{\mu_1},t)\leq f(\tilde{\mu_1},t) $$ ?
Remark: recall that $\sum_{i=1}^N \tilde{\mu}_i=1$ and $\sum_{i=1}^N \bar{\mu}_i=1$.
No. Take $N=3, \bar{\mu} = (\frac{1}{2},\frac{1}{2},0), \tilde{\mu} = (\frac{1}{2},0,\frac{1}{2}), t_1 = 1, t_2 = 0, t_3 = 1$. Then, $f(\bar{\mu_1},t) = \frac{\frac{1}{2}}{\frac{1}{2}} = 1$ while $f(\tilde{\mu_1},t) = \frac{\frac{1}{2}}{\frac{1}{2}+\frac{1}{2}} = \frac{1}{2}$.