I'll put my comment as an answer. A butterfly is a combination of a straddle and a strangle. Let's assume the straddle is ATM and the strangle OTM. The price of an option $V$ is bounded $0 \leq V \leq S$. It can't exceed the value of the underlying.
The worst that can happen for an option holder is that the underlying doesn't move. Assume that the underlying doesn't move at all in which case the OTM options are rendered worthless, $OTM_C=OTM_P=0$. Also assume that the ATM option has maximum value $S$. That yields a total cost of $ATM_C+ATM_P-OTM_C-OTM_P=2S$.
I know this is making some distributional assumptions but you can verify the thought process using Black Scholes. I.e. put the vol of the OTM call and put equal to zero (or an extremely small value). This is because $N(d_1)$ and $N(d_2)$ both become zero since the $d_1$ and $d_2$ values become extremely negative (in case of the OTM call). Repeat the exercise and put the vol for the ATM to a very large value, resulting in an option price equal to that of the underlying. See below example from a pricer:
If the OTM wings have any value larger than zero, the cost of the strategy will cheapen and be less than $2S$. Similarly, if the ATM options are worth less than $S$ the total cost will become cheaper.