I have that:

$\displaystyle \prod_{i=1}^n \big (1-y_i \big ) = d$, where $\displaystyle 0 \leq y <1 $

I can use the following to decompose the products:

Lets say for n=10:

$\displaystyle \prod_{i}^{10} (1-y_i) = 1 - \sum_{i}^n y_{i} + \sum_{i,j}y_{i}y_{j} - \sum_{i,j,k}y_{i}y_{j}y_{k} + \sum_{i,j,k,l}y_{i}y_{j}y_{k}y_{l} - \sum_{i,j,k,l,m}y_{i}y_{j}y_{k}y_{l}y_m + \sum_{i,j,k,l,m,p} y_{i}y_{j}y_{k}y_{l}y_m y_p - \sum_{i,j,k,l,m,p,r} y_{i}y_{j}y_{k}y_{l}y_m y_p y_r + \sum_{i,j,k,l,m,p,r,s} y_{i}y_{j}y_{k}y_{l}y_m y_p y_ry_s - \sum_{i,j,k,l,m,p,r,s,u} y_{i}y_{j}y_{k}y_{l}y_m y_p y_ry_sy_u + y_{1}y_{2}y_{3}y_{4}y_5 y_6 y_7 y_8y_9y_{10} = d$

I have empirically observed that all of the following inequalities hold. But I don't know why? Can anybody prove this?

$\displaystyle 1 - \sum_{i}^n y_{i} + \sum_{i,j}y_{i}y_{j} \geq d$

$\displaystyle 1 - \sum_{i}^n y_{i} + \sum_{i,j}y_{i}y_{j} - \sum_{i,j,k}y_{i}y_{j}y_{k} + \sum_{i,j,k,l}y_{i}y_{j}y_{k}y_{l} \geq d$

$\displaystyle 1 - \sum_{i}^n y_{i} + \sum_{i,j}y_{i}y_{j} - \sum_{i,j,k}y_{i}y_{j}y_{k} + \sum_{i,j,k,l}y_{i}y_{j}y_{k}y_{l} - \sum_{i,j,k,l,m}y_{i}y_{j}y_{k}y_{l}y_m + \sum_{i,j,k,l,m,p} y_{i}y_{j}y_{k}y_{l}y_m y_p \geq d$

$\displaystyle 1 - \sum_{i}^n y_{i} + \sum_{i,j}y_{i}y_{j} - \sum_{i,j,k}y_{i}y_{j}y_{k} + \sum_{i,j,k,l}y_{i}y_{j}y_{k}y_{l} - \sum_{i,j,k,l,m}y_{i}y_{j}y_{k}y_{l}y_m + \sum_{i,j,k,l,m,p} y_{i}y_{j}y_{k}y_{l}y_m y_p - \sum_{i,j,k,l,m,p,r} y_{i}y_{j}y_{k}y_{l}y_m y_p y_r + \sum_{i,j,k,l,m,p} y_{i}y_{j}y_{k}y_{l}y_m y_p y_ry_s \geq d $

$\displaystyle 1 - \sum_{i}^n y_{i} + \sum_{i,j}y_{i}y_{j} - \sum_{i,j,k}y_{i}y_{j}y_{k} + \sum_{i,j,k,l}y_{i}y_{j}y_{k}y_{l} - \sum_{i,j,k,l,m}y_{i}y_{j}y_{k}y_{l}y_m + \sum_{i,j,k,l,m,p} y_{i}y_{j}y_{k}y_{l}y_m y_p - \sum_{i,j,k,l,m,p,r} y_{i}y_{j}y_{k}y_{l}y_m y_p y_r + \sum_{i,j,k,l,m,p,r,s} y_{i}y_{j}y_{k}y_{l}y_m y_p y_ry_s - \sum_{i,j,k,l,m,p,r,s,u} y_{i}y_{j}y_{k}y_{l}y_m y_p y_ry_sy_u + y_{1}y_{2}y_{3}y_{4}y_5 y_6 y_7 y_8y_9y_{10} \geq d $

I did experiments in matlab for up to n = 50 and all of them always hold.