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Without additional assumptions this can not be true. If you fix the value of $T:=R^{1/\sigma-1}$ and let $\sigma$ tend to $1$, $B(n_1)$ tends to $(N-n_1)(p+(1-p)T)$ (as this is the expectation of $\xi_1+\ldots+\xi_{N-n_1}$, where $\xi_i$ are i.i.d. taking the values 1 or $T$ with probabilities $p$ and $1-p$ respectively). Then, since $T>1$, the inequality holds with the opposite sign.