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Take 2 sets of real numbers:

  • $x_1, x_2, \dots, x_n$ and $y_1, y_2, \dots, y_m$ such that $\prod\limits_{1 \le i \le n} x_i > \prod\limits_{1 \le j \le m} y_j$.

Let $k$ be any positive real number.

Does it necessarily follow that $\prod\limits_{1 \le i \le n} (x_i+k) > \prod\limits_{1 \le j \le m} (y_j +k)$

If the question were related to addition, then the generalization would apply: $$\sum\limits_{i=1}^n\left(x_i + k\right) > \sum\limits_{j=1}^m \left(y_j + k\right)$$

Intuitively, adding a positive to each real $x_i$ and each real $y_j$ should increase the product so the question relates to how much it increases each product.

For example, if I choose $x_1 = 10, x_2, = 11$ and $y_1 =1, y_2 = 2$ and $k=1$ it is clear that $110 > 2$ and $132 > 6$.

It seems to me that the answer is yes. Am I correct? If yes, how does one prove this? If no, what is the argument against?

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    $\begingroup$ Consider difference of these products as a polynomial of $k$: $d(k) = \prod(x_i + k) - \prod(y_i - k)$ – it has degree $n$, and the only thing we know about it is that $d(0) > 0$ :) $\endgroup$ – Anton Grudkin Sep 29 '20 at 4:17
  • $\begingroup$ Actually, degree at most $n-1$, and possibly less. For instance, $(13,18)$ and $(6,25)$ yield a difference of degree $0$ (i.e. a constant), because $13+18=6+25$. For $n=2$, there will be a sign change if $a+b-c-d$ and $ab-cd$ don't have the same sign. This happens, but not that much: among all $10^8$ values of $(a,b,c,d)$ between $1$ and $100$, around $7.8\%$ satisfiy this condition. $\endgroup$ – Jean-Claude Arbaut Dec 30 '20 at 23:40
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No, it isn't always true. For example, let $n = 2$, $x_1 = x_2 = 3$, $y_1 = 1$, $y_2 = 8$ and $k = 1$. Then we have

$$\prod_{i=1}^{2}x_i = 3(3) = 9 \gt \prod_{j=1}^{2}y_j = 1(8) = 8 \tag{1}\label{eq1A}$$

but

$$\prod_{i=1}^{2}(x_i + k) = 4(4) = 16 \lt \prod_{j=1}^{2}(y_j + k) = 2(9) = 18 \tag{2}\label{eq2A}$$

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  • $\begingroup$ Thanks, John. Great example. $\endgroup$ – Larry Freeman Sep 29 '20 at 3:34
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    $\begingroup$ @LarryFreeman You're welcome. For these types of questions, I find trying "extreme" type cases, i.e., where one set has values being the same while the other has values about as far apart as possible instead, can be used to help do a relatively quick check on whether or not there's a counter-example, or at least give an idea of likely there is to be one. $\endgroup$ – John Omielan Sep 29 '20 at 3:36
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Irrelevant at this point, but I started this answer before the question was answered, so I might as well put it up.

Probably the easiest way to understand this is to take logs. Since everything is positive, and the logarithm is increasing, we have that $$ \prod\limits_{i=1}^n\left(x_i + k\right) > \prod\limits_{j=1}^n \left(y_j + k\right)$$ if and only if $$\log\left(\prod\limits_{i=1}^n\left(x_i + k\right)\right) > \log\left(\prod\limits_{j=1}^n \left(y_j + k\right)\right)$$ which in turn is equivalent to $$\sum\limits_{j=1}^n\log\left(x_j + k\right) > \sum\limits_{j=1}^n\log\left(y_j + k\right) $$

But this you can see is actually not true, because of the behavior of the natural log. It has sort of diminishing returns, right? So, one idea is to construct a scheme where the LHS has inputs to the log that are all too big, so the collective impact of the greater increases on the right is enough to make the difference. Notice also that if there doesn't have to be the same number of these numbers, this is quite easy. But, with the idea in mind, let's suppose that the RHS has a lot of numbers that are quite small.

So, let's take $x$ to be the sequence $1, 1, \ldots, 1$ with 10 elements. Then we take $y$ to be the sequence$10^8$, then $.1$ 9 times. Both of these sequences have 10 elements and the product of $x$ is 1 which is bigger than that of $y$, which is $\frac{1}{10}$. But if I add 1 to $x$ the product becomes just $1024$, which when I do it to $y$ it becomes $(10^8+1)(1.1)^9$, which is larger.

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    $\begingroup$ Thanks, Cade. I appreciate your explanation. This gives me more to think about! :-) $\endgroup$ – Larry Freeman Sep 29 '20 at 3:43
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A counterexample is given by $m=n=2,\; x_1=x_2=-1,\; y_1=y_2=0,\; k=1.$

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