Upper bound for product of exponents

Question

From here we have the bound

$$\left(1-\frac1N\right)^N\leq e^{-1}$$

where $N$ is a positive integer.

Written another way, it is

$$\left(1-\frac1{N_1}\right)^{k_1}\left(1-\frac1{N_2}\right)^{k_2}\ldots \left(1-\frac1{N_r}\right)^{k_r}\leq e^{-1}$$ where $N_1=N_2=\ldots=N_r=N$ and $\dfrac{k_1}{N_1}+\dfrac{k_2}{N_2}+\ldots+\dfrac{k_r}{N_r}=1$.

Is it true in general that

$$\left(1-\frac1{N_1}\right)^{k_1}\left(1-\frac1{N_2}\right)^{k_2}\ldots \left(1-\frac1{N_r}\right)^{k_r}\leq e^{-1}$$

for positive integers $N_1,N_2,\ldots N_r,k_1,k_2,\ldots k_r$ such that $\dfrac{k_1}{N_1}+\dfrac{k_2}{N_2}+\ldots+\dfrac{k_r}{N_r}=1$? If not, what upper bound can we find for the left-hand expression?

Answer 1

In full generality, for every nonnegative $n$ and every $a\lt1$, $$\left(1-a\right)^n\leqslant\mathrm e^{-an}. $$ Using this for $a=1/N_i$ and $n=k_i$ yields $\left(1-1/N_i\right)^{k_i}\leqslant\mathrm e^{-k_i/N_i}$ hence $$ \prod_i\left(1-\frac1{N_i}\right)^{k_i}\leqslant\exp\left(-\sum_i\frac{k_i}{N_i}\right). $$

Answer 2

I think I just got it. Following the proof in the link:

$$\left(1-\dfrac{1}{N_1}\right)\leq e^{-1/N_1}$$

so $$\left(1-\dfrac{1}{N_1}\right)^{k_1}\leq e^{-k_1/N_1}$$

Therefore, $$\left(1-\frac1{N_1}\right)^{k_1}\left(1-\frac1{N_2}\right)^{k_2}\ldots \left(1-\frac1{N_r}\right)^{k_r}\leq e^{-(k_1/N_1+k_2/N_2+\ldots+k_r/N_r)}=e^{-1}.$$