Prove $ P_{n+1} + _{n+2} ≤ _1\times_2\times\cdots\times_n$ for $n\geq3$. $P_n$ is the $n$-th prime.

Question

As part of one of my assignment in CS degree, I have to prove this question:

$P_n$ is the $n$-th prime. Prove $$P_{n+1} + P_{n+2} \leq P_1\times P_2\times\cdots\times P_n$$ for $n\geq3$.

I was trying to apply Bonse's inequality, which indicate $$P_{n+1}^2\lt P_1\times P_2\times\cdots\times P_n$$ for $n\geq4$, but with no successes. Any help will be grateful.

Answer 1

Bonse's inequality is true for $n\geq4$ (it is widely accepted that $1$ is not a prime, thus $p_1=2$). Also for $n\geq4 \Rightarrow p_n>3$. What we have is $$p_1\cdot p_2\cdot ... \cdot p_n > p_{n+1}^2>3p_{n+1}=2p_{n+1}+p_{n+1} > ...$$ Now, the key point is Bertrand's postulate, $2p_{n+1}>p_{n+2}$, thus $$...>p_{n+2}+p_{n+1}$$ You will have to check $n=3$ case manually.

Answer 2

Apply Bonse at $P_{n+1}$ and at $P_{n+2}$ and get $$P_{n+1}^2+P_{n+2}^2 \le (P_1 \cdots P_n) \cdot (1+P_{n+1}). \tag{1}$$

A re-do of finish (wrong before). Let $Q$ be the prime product which is the first factor on the right of $(1)$. Also let $p=P_{n+1},q=P_{n+2}.$ Then dividing $(1)$ by $(1+p)$ we have $$\frac{p^2+q^2}{1+p} \le Q. \tag{2}$$ So it will be enough to show $p+q$ is bounded above by the left side of $(2).$ That is, $p+q+pq+q^2 \le p^2+q^2,$ or $$ p+q+pq \le q^2.$$ Since $p,q$ are odd primes, if we put $q=x$ then $p \le x-2$ and the left side is at most (replacing $p$ by $q-2$) $x^2-2,$ bounded above as desired by $q^2=x^2.$

Answer 3

Bertrand's postulate alone suffices.

$p_{n+1}<2p_n,\ p_{n+2}<2p_{n+1}<4p_n$

$p_{n+1}+p_{n+2}<6p_n$

$p_1\cdot p_2=6$, so $p_n\# \ge 6p_n$ for $n\ge 3$

Ergo $p_{n+1}+p_{n+2}<p_n\# $