How do you show that, if $p_n$ denotes the $n$th prime, $n > 3$, then
$$p_n < p_1 + p_2 +\cdots + p_{n-1}$$
using the Bertrand conjecture and induction?
Here is what I've come up with, but I'm not sure if it's correct:
Let $n=4$. Then, $p_4 = 7$, so we have:
$$7 < 2+3+5 = 10.$$
Assume that the statement is true for $n$ so that $p_n < p_1 + p_2 +\cdots + p_{n-1}$.
We want to prove that this is true for $n+1$.
$$p_{n+1} < p_1 + p_2 +\dotsc + p_{n-1} + p_n.$$
But we assumed that $p_n < p_1 + p_2 +\cdots + p_{n-1}$, so we have:
$$p_{n+1} < p_1 + p_2 +\cdots + p_{n-1} + p_n < p_n + p_n = 2p_n.$$
We know that $p_n < p_{n+1}$, so we have:
$p_n < p_{n+1} < 2p_n$, which is true by the Bertrand conjecture. $\blacksquare$