Estonian Math Competitions

First notice that the $n$-th prime $p_n$ satisfies the inequality $p_n \ge 2n - 1$. Indeed, the claim holds for the first prime $p_1 = 2$. Since all other primes are odd and there is exactly $n-1$ odd numbers between $2$ and $2n$, there are at most $n$ prime numbers less or equal to $2n-1$, hence $p_n \ge 2n - 1$.

Now consider the sum of the $n$ first primes $P = p_1 + p_2 + \dots + p_n$. Since $p_k \ge 2k - 1$ for any $k$, and additionally $p_1 = 2 > 1$, the sum $P$ is strictly greater than the sum of $n$ first odd numbers $S = 1 + 3 + \dots + (2n-1) = (1^2 - 0^2) + (2^2 - 1^2) + \dots + (n^2 - (n-1)^2) = n^2$. So $P > S = n^2$.