Is $2p_ip_j\lt p_{j+1}^2$ for $i\lt j$? Initially, I was wondering is the following statement true or not-

Let $p_k$ be the $k^{\text{th}}$ prime.
Then, $2p_ip_j\lt p_{j+1}^2$, where $i\le j$

But then I realized that the case $i=j$ has obvious counter-examples like-
$2 \times 5^2 \gt 7^2$
So I excluded the possibility of $i=j$, then the statement becomes

Let $p_k$ be the $k^{\text{th}}$ prime.
Then, $2p_ip_j\lt p_{j+1}^2$, where $i\lt j$.

I cannot find any counter-examples for this. So I thought about using Bertrand's Postulate which states
$2p_j\gt p_{j+1}$ and the fact that $\frac{p_i}{p_{j+1}}\lt 1$ since $i\lt j+1$.
So, $\frac{p_i}{p_{j+1}}\lt 1$
$\implies -\frac{p_i}{p_{j+1}}>-1$
Multiplying this with the statement of Bertrand's Postulate gives us
$-\frac{2p_ip_j}{p_{j+1}}>-p_{j+1}$
$\implies 2p_ip_j<p_{j+1}^2$
Obviously, we can't just multiply inequalities, especially when dealing with negative numbers and not only that but this line of reasoning also implies the case $i=j$ since $j\lt j+1$ which I just disproved.
To validate my reasoning I was trying to investigate how to multiply inequalities when dealing with negative numbers but that quickly became too complicated for me to work with.
So my question is-
Is it possible to validate my reasoning so that it works? If not, I would like to have a proof of the statement.

Try a few examples before attempting to prove anything.
2 x 13 x 17 > 361
2 x 17 x 19 > 529
2 x 19 x 23 > 841
2 x 23 x 29 > 961
2 x 29 x 31 > 1369
2 x 31 x 37 > 1681
2 x 37 x 41 > 1849
2 x 41 x 43 > 2209
2 x 43 x 47 > 2809
2 x 47 x 53 > 3481
2 x 53 x 59 > 3721
2 x 59 x 61 > 4489
2 x 61 x 67 > 5041
2 x 67 x 71 > 5329
2 x 71 x 73 > 6241
2 x 73 x 79 > 6889
2 x 79 x 83 > 7921


Your attempt to use Bertrand's postulate leads to the opposite direction. Feel it with your heart! Bertrand's postulate states the closeness of primes -- the next prime will never be further then twice this prime. However, your conjecture states the apartness of primes -- twice the product of smaller primes will never reach the square of a bigger prime.
From this, we can conclude that, if you are to prove your conjecture (which you can't, but we can learn from it), Bertrand's postulate will never, ever help you! It will always push you towards the other direction. Unless..
Unless, you come across a situation where an inequality is "flipped" by a constraint. For instance, given $a+b+c + abc = 10$, and suppose you have the inequality $a+b+c > 3$, which states that $a,b,c$ are big. Then you can get the inequality $abc < 7$, stating that $a,b,c$ are small. This is an example of how inequalities may be "flipped".
