Comparing two expressions involving exponents

Question

I was working on a probabilty question where I was trying to compute probabilty of error in two events. I was able to obtain the following:

$\mathbb{P}(err_A) \leq n_1e^{-Tp(1-p)^{k_1}} + n_2e^{-Tp(1-p)^{k_2}}$ and $\mathbb{P}(err_B) \leq(n_1+n_2)e^{-Tp(1-p)^{k_1+k_2}}$

I know that $n_1,n_2, T, k_1,k_2$ are positive integers with $n_1,n_2 \gt T >>k_1,k_2 $. Also $0\leq p\leq 1$.

I was trying to compare the 2 RHS bounds to see which one is tighter. Specifically, which is greater $n_1e^{-Tp(1-p)^{k_1}} + n_2e^{-Tp(1-p)^{k_2}}$ or $(n_1+n_2)e^{-Tp(1-p)^{k_1+k_2}}$.

Answer 1

We have $0<e^{-Tp(1-p)^{k_i}}<1$ and $0 \le (1-p)^{k_i} \le 1$, $i=1,2$.

For $0<a<1$ and $0\le b\le 1$, we have $a^b \ge a$, with equality at $b=1$. Hence

\begin{align}(n_1+n_2)e^{-Tp(1-p)^{k_1+k_2}}&=n_1(e^{-Tp(1-p)^{k_1}})^{(1-p)^{k_2}}+n_2(e^{-Tp(1-p)^{k_2}})^{(1-p)^{k_1}} \\&\ge n_1e^{-Tp(1-p)^{k_1}}+n_2e^{-Tp(1-p)^{k_2}}\end{align}

with equality at $(1-p)^{k_i} = 1$, that is, if and only if $p=0$.