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.