pr.probability - A probabilistic inequality

I think it's true.
My reasoning goes like this (and I kept making mistakes with the algebra/arithmetic so check carefully)

$P(S_6 lt 3 delta) le P(hbox{exactly 4 }X_ihbox{s less than }delta)+P(hbox{exactly 5 }X_ihbox{s less than }delta)+P(hbox{all 6 }X_ihbox{s less than }delta)$. So if we let $p=P(X_1 lt delta)$, then the right hand part of this inequality is

$(p^4)(10p^2-24p+15)$. I then plugged $2p-(p^4)(10p^2-24p+15)$ into R and got a function that looks always non-negetive between 0 and 1 (shouldn't be too bad to show this by derivatives but I'm lazy and have computer power).