I want to show that $a+b+c \leq 2 + 2abc$ for $0 \leq a,b,c \leq 1$.
I think this inequality is true, but I don't know how to show it properly. I have tried substituting $a=b=c=t$, which gives $3t \leq 2+2t^3$, which seems to be true.
I tried $a=b=(1-t)$, so we have $2-2t+c \leq 2 + 2(1-t)^2 c$, which I work out to be true also.
I can show that the easier inequality $a+b+c \leq 2 + 2ab$ (assuming that $a<b<c$).