the probability of a decision error in a 3-bit majority
encoding is
perror = 3p^2 - 2p^3
where p is the probability of a single bit flipping. Show perror < p, if
p < 1/2. (NOTE: I want a formal proof. That means you can?t simply
plug several values in for p and expect that to be the correct answer.)

p>0
3p^2 < p if p< 1/3 as 3*p<1, implying 3p*p < p
as p is non-negative, 2p^3 is also non-negative.
this, 3p^2 - 2p^3 < 3p^2 < p

have fun with 1/3<p<1/2

Thank you for the comment. I think I've solved it, but thanks again.