# A number theory question about a “double infimum”

Let $x_1,x_2,x_3,\ldots,x_S$ be numbers with $x_i>-1$ for all $i$ and $x_k<0$ for some $k$.

How can one show that $$\inf_{s\in[1,S]}\inf_{t\in[1,s]}\prod_{i=t}^s (1+\frac{1}{2}x_i) < \inf_{s\in[1,S]}\inf_{t\in[1,s]}\prod_{i=t}^s (1+\frac{1}{4}x_i)$$ This seems to instinctively be obvious, because the "most destructive path" surely must be a bit less destructive when we reduce the "destruction" from 1/2 to 1/4, but I'm not sure how to formalize this thought.

Update: Try also to generalize this for not just 1/2 and 1/4, but for any number $q$ and any other number $p<q$, with $0<p,q<1$.

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Actually, the answer (including the $(p,q)$-case) is already there. –  Did Aug 15 '12 at 10:30

For convenience, define $f(s,t) = \prod_{t=i}^s \left(1 + \frac{1}{4} x_i\right)$ and $g(s,t) = \prod_{t=i}^s \left(1 + \frac{1}{2} x_i\right)$. Note that since $x_i \leq 0$, both $f(s,t)$ and $g(s,t)$ are monotonically decreasing in $s$ and monotonically increasing in $t$. Therefore, $$f(S,1) = \inf_{s \in [1,S]} \inf_{t \in [1,s]} f(s,t) \quad \text{and} \quad g(S,1) = \inf_{s \in [1,S]} \inf_{t \in [1,s]} g(s,t)\enspace.$$ Since $x_j \leq 0$ for all $j$, we have $$\prod_{i \in A} \left( 1 + \frac{1}{4} x_i \right) \geq \prod_{i \in A} \left( 1 + \frac{1}{2} x_i \right) \enspace,$$ for any set $A \subset ( \mathbb{N} \cap [1,S] )$. To get a strict inequality, pick any $x_k$ such that $x_k < 0$ (and $k \in [1,S]$) and define $A = ((\mathbb{N} \cap [1,S]) \setminus \{x_k\})$. Clearly, \begin{align*} f(S,1) = \prod_{i=1}^S \left( 1 + \frac{1}{4} x_i \right) & = \left( 1 + \frac{1}{4} x_k \right) \prod_{i \neq k}^S \left( 1 + \frac{1}{4} x_i \right) \\ & \geq \left( 1 + \frac{1}{4} x_k \right) \prod_{i \neq k}^S \left( 1 + \frac{1}{2} x_i \right) \\ & > \left( 1 + \frac{1}{2} x_k \right) \prod_{i \neq k}^S \left( 1 + \frac{1}{2} x_i \right) = \prod_{i=1}^S \left( 1 + \frac{1}{4} x_i \right) = g(S,1) \enspace. \end{align*} And we are done.

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I never mentioned that $x_j\le 0$ for all $j$..., just for some $j$. –  godel68 Aug 1 '12 at 23:10
Very true. I'm sorry, I misread. –  MLS Aug 1 '12 at 23:34

The observation that $1+\frac{1}{2}x \le \left(1 + \frac{1}{4}x\right)^2$ for all $x$, and that the inequality is strict whenever $x\neq 0$, is all you need. Then $$\prod_{i=t}^{s}\left(1 + \frac{1}{2}x_i\right) < \left(\prod_{i=t}^{s}\left(1 + \frac{1}{4}x_i\right)\right)^2$$ for any $t \le s$ as long as $x_i \neq 0$ for some $t\le i \le s$. The double infinimum of the latter product over $s$ and $t$ must be less than $1$, since we're given that at least one $x_i$ is negative, and hence squaring it makes it even smaller.

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Do you know how the result could be made more general, using not just 1/2 vs. 1/4, but rather any number $q$ vs. any other number $p<q$? (with $0<p,q<1$). –  godel68 Aug 1 '12 at 23:10