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I have noticed the following curiosity for a product of integers. Given an ordered (decreasing) sequence of strictly positive integers $(a_i)_{i=1 \ldots n}$, that is to say, such that:

$$\forall\ i\ a_{i+1} \ge a_i,$$

I've noticed in concrete examples this surprising relation: $$\Pi_{i=1}^n a_i \le \Pi_{i=1, i \neq j, i \neq k}^n a_i (a_j-1)(a_k+1)\ \forall\ j \le k$$

In other words, for two positive integers $a$ and $b$ with $b \ge a$ it seems that $ab \le (b-1)(a+1)$.

Is this relation always true ? Where can I find content related to this thing ?

Thank you.

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You can expand $(b-1)(a+1) = ba +b - a - 1$, so If $b > a+1$, it is larger than $ab$, and if $b \leqslant a$, it is smaller. – Daniel Fischer Oct 18 '13 at 19:28

$$(b-1)(a + 1) = ba + b - a - 1$$

So you'll want $b > a$ for this to be bigger than $ab$.

For example where it fails without the strict inequality, just notice $$(2-1)(2 + 1) = 3 < 2\cdot 2 = 4$$

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Exact, thank you. – Dingo13 Oct 18 '13 at 19:31

This ties nicely into a fun geometry exercise. Given a fixed parameter, which rectangle will have the greatest area? Well, given two sides $a$ and $b$, $b \geq a$, the perimeter is $2(a+b)$ and the area is $ba$. Now if we increase $a$ by $\epsilon$ and decrease b by the same, the perimeter will be the same but the area will become $ba + \epsilon (b - a) - \epsilon^2$. Take the derivative wrt $\epsilon$ and we get the greatest area when $(b-a)/2 = \epsilon$. And hey, $b - (b-a)/2 = a+(b-a)/2 = (a+b)/2$. The rectangle has the greatest area when the sides are equal, so a square.

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