Inequality involving prime numbers

Question

If $p_k$ denotes the $k$ th prime then show that $$p_k \cdot p_{k+1} > p_{k+2}.$$ $$$$ I think that Bertrand's pastulate and Bonse's inequality could be helpful but I exactly don't know how to use them?

Answer 1

Hint: $p_k\cdot p_{k+1}\geq 2p_{k+1}$

Answer 2

You can prove this as follows $P_{k+1}>P_k \implies P_k\cdot P_{k+1}>P_k^2$

but we know there is always a prime between $a$ and $2a$[Bertrand's Postulate]. hence for $P_k>3$ $\implies P_k^2> P_{k+2}$

combining the 2 we get

$P_k\cdot P_{k+1}>P_{k+2}$