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

$P(S_6 \lt 3 \delta) \le P(\hbox{exactly 4 }X_i\hbox{s less than }\delta)+P(\hbox{exactly 5 }X_i\hbox{s less than }\delta)+P(\hbox{all 6 }X_i\hbox{s less than }\delta)$. 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).