TITLE: prove or disprove $P_{i}\cdot P_{k+1-i}>P_{k+1}$
QUESTION [10 upvotes]: let $p_{k}$ is  the $k$-th prime,
prove or disprove
$$P_{i}\cdot P_{k+1-i}>P_{k+1},1\le i\le k$$
I find this
$$P_{1}=2,P_{2}=3,P_{3}=5,P_{4}=7,P_{5}=11,P_{6}=13,P_{7}=17,P_{8}=19,\cdots$$
then
$$P_{1}P_{2}=6>p_{3}=5$$
$$P_{1}P_{3}=10>P_{4}=7$$
$$P_{1}P_{4}=14>P_{5}=11$$
$$\cdots\cdots$$
$$P_{2}P_{3}=15>P_{5}=11$$
$$P_{2}P_{4}=21>P_{6}=13$$
$$P_{2}P_{5}=33>P_{7}=17$$
$$\cdots\cdots$$
and so on
and I  find this :Is it true that $p_{n}+p_{n+1}>p_{n+2}$ for all $n\geq 2\ ?$
and What's the history of the result that $p_{n+1} < p_n^2$, and how difficult is the proof?
But also I can't prove (or disprove )this equality,
I found this problem is strong this Bertrand theorem
Thank you  for you help,

REPLY [5 votes]: Let's prove that for integers $b \geq a\geq 1$, we have $P_aP_b > P_{a+b}$ (this is slightly stronger than the original statement, since it allows for $a=b$).
There are only ten pairs $(a,b)$ with $b<5$ and it's easy to verify that the statement is true for all of them. Thus, from now on, we'll assume $b\geq 5$.
Weakening the bounds from Wikipedia a little, we get the following inequality, valid for $n\geq 5$:
$$n\ln n < P_n < 1.37n\ln n$$
Applying the upper bound to $P_{a+b}$ and using the fact $2b \geq a+b$ yields 
$$b\log (6.7b^{2.74}) > 2.74b\log(2b) \geq 1.37(a+b)\log(a+b) > P_{a+b}$$
The lower bound applied to $P_b$ gives us
$$P_aP_b > P_a b\log b=b\log b^{P_a}$$
For $P_a\geq 5$, we have $b^{P_a}\geq b^5>6.7b^{2.74}$, so we can combine the two inequalities to obtain $P_aP_b>P_{a+b}$. The same is true for $P_a=3$ and $b\geq 1504$. The remaining cases with $P_a=3$ and $2\leq b<1504$ can be checked by a computer (or by hand, if you prefer so; it turns out to be quite easy if you have the list of first ~$1500$ primes by hand).
Finally, $P_a=2$ corresponds to Bertrand's postulate, $2P_b > P_{b+1}$. Q.E.D.