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?
On
$\alpha(1-\beta)\times\beta(1-\gamma)\times\gamma(1-\alpha) = \alpha(1-\alpha)\times\beta(1-\beta)\times\gamma(1-\gamma)=X$
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 $1/4)$
On
Because by AM-GM $$\prod_{cyc}\alpha(1-\beta)=\prod_{cyc}\alpha(1-\alpha)\leq\prod_{cyc}\left(\frac{\alpha+1-\alpha}{2}\right)^2=\frac{1}{64}$$ and we'll got a contradiction if $$\alpha(1-\beta)>\frac{1}{4},$$ $$\beta(1-\gamma)>\frac{1}{4}$$ and $$\gamma(1-\alpha)>\frac{1}{4},$$ which gives $$\prod_{cyc}\alpha(1-\beta)>\frac{1}{64}.$$
On
Assume not. Let $s:=\min\{\alpha,\beta,\gamma, 1-\alpha,1-\beta,1-\gamma\}$, so clearly $0<s\le \frac12$. Then $$\frac14<\alpha(1-\beta)\le\alpha(1-s),\\ \frac14<(1-\alpha)\gamma\le (1-\alpha)(1-s),$$ and likewise $$ \frac14<\beta(1-s),\quad\frac14<(1-\beta)(1-s),\quad\frac14<\gamma(1-s),\quad\frac14<(1-\gamma)(1-s)$$so that also
$$ \frac14<s(1-s)=\frac14-\left(s-\frac12\right)^2$$ conradiction.
WLOG let $\alpha \ge \gamma$. Then we have:
$$\gamma(1-\alpha) \le \alpha(1-\alpha) = \alpha - \alpha^2 = \frac 14 - \left(\frac 12 - \alpha\right)^2 \le \frac 14$$