It is possible to "compact" this notation? I need some help with notations. Suppose that the inequality
$$ 0\leq s_i\leq p_i p_j^{\ast} \quad\mbox{ for } i, j =1, 2 \mbox{ and } i\neq j$$
holds, where $p^{\ast}$ denote the Sobolev critical exponent. I need to write in "compact form" (I mean as in the inequality above) the following inequalities
$$ 1<s_3<p_1^{\ast}, \quad 1<\frac{s_1 s_3}{s_3-1}<p_2^{\ast}, \quad 1< s_4 <p_2^{\ast},\quad 1<\frac{s_2 s_4}{s_4-1}<p_1^{\ast}.$$
Could anyone please help me?
Thank you in advance!
EDIT: for $i, j = 1, 2, 3$ with $i\neq j$ it is $0\leq s_i\leq p_i\min(p_j^{\ast})$, I need to compact the expression:
$$1 <s_4 <p_1^{\ast}, \quad 1< s_5<p_2^{\ast}, \quad 1< s_6 <p_3^{\ast}$$
and
$$1<\frac{s_1 s_4}{s_4 -1}<\min(p_2^{\ast}, p_3^{\ast},\quad 1<\frac{s_2 s_5}{s_5-1}<\min(p_1^{\ast}, p_3^{\ast}), \quad 1<\frac{s_3 s_6}{s_6 -1}<\min(p_1^{\ast}, p_2^{\ast}).$$
 A: Assuming the following relationship is valid for $j=4$:
\begin{align*}
&1 <s_5<p_1^{\ast}, \quad 1< s_6<p_2^{\ast}, \quad 1< s_7 <p_3^{\ast}, \quad 1< s_8 <p_4^{\ast}\\
\\
&1<\frac{s_1 s_5}{s_5 -1}<\min\{p_2^{\ast}, p_3^{\ast}, p_4^{\ast}\},\quad
1<\frac{s_2 s_6}{s_6-1}<\min\{p_1^{\ast}, p_3^{\ast}, p_4^{\ast}\}\\
\\
&1<\frac{s_3 s_7}{s_7-1}<\min\{p_1^{\ast}, p_2^{\ast}, p_4^{\ast}\},\quad
1<\frac{s_4 s_8}{s_8-1}<\min\{p_1^{\ast}, p_2^{\ast}, p_3^{\ast}\}\\
\end{align*}

We might conclude for general $j\geq1$:
\begin{align*}
&1 <s_{j+1}<p_1^{\ast}, \quad 1< s_{j+2}<p_2^{\ast}, \quad \cdots, \quad 1< s_{2j} <p_j^{\ast}\\
\\
&1<\frac{s_1 s_{j+1}}{s_{j+1} -1}<\min\{p_2^{\ast},p_3^{\ast},\ldots,  p_j^{\ast}\},\quad
1<\frac{s_2 s_{j+2}}{s_{j+2}-1}<\min\{p_1^{\ast}, p_3^{\ast},\ldots, p_j^{\ast}\}\\
&\qquad\qquad\cdots\\
&1<\frac{s_j s_{2j}}{s_{2j}-1}<\min\{p_1^{\ast}, p_2^{\ast}, \ldots,p_{j-1}^{\ast}\}\\
\end{align*}

or more compactly for general $j\geq 1$ and $1\leq k\leq j$:

\begin{align*}
&\color{blue}{1 <s_{j+k}<p_k^{\ast},} \\
\\&\color{blue}{1<\frac{s_k s_{j+k}}{s_{j+k} -1}<\min\{p_1^{\ast},\ldots,  p_j^{\ast}\}\setminus\{p_k^{\ast}\}}
\end{align*}

