Proving at least one number $\le 1/4$ How to prove that at least one of $\alpha(1-\beta),\beta(1-\gamma),\gamma(1-\alpha)$ is lesser than $1/4$ given that $\alpha,\beta,\gamma\in(0,1)$

The original problem was proving at least one triangle (except the middle one) formed by dividing an equilateral triangle into four triangles is at least 4 times smaller than the original which I simplified to above. Don't know how to proceed, any hints?
WLOG let $\alpha \ge \gamma$. Then we have:

$$\gamma(1-\alpha) \le \alpha(1-\alpha) = \alpha - \alpha^2 = \frac 14 - \left(\frac 12 - \alpha\right)^2 \le \frac 14$$
$\alpha(1-\beta)\times\beta(1-\gamma)\times\gamma(1-\alpha) = \alpha(1-\alpha)\times\beta(1-\beta)\times\gamma(1-\gamma)=X$

Since $\alpha,\beta,\gamma$ are independent we can find maximum value of $\alpha(1-\alpha)=1/4$ at $\alpha=1/2$. Similarly for other two.
Therefore $X \leq 1/16$. 

If product of $\alpha(1-\beta),\beta(1-\gamma),\gamma(1-\alpha)$ is $\leq 1/16$, atleast one of $\alpha(1-\beta),\beta(1-\gamma),\gamma(1-\alpha)$ is $\leq1/4$ and atleast one of $\alpha(1-\beta),\beta(1-\gamma),\gamma(1-\alpha)$ is $\geq1/4$. $($Since at $\alpha=\beta=\gamma=1/2$ all three are $1/4)$