1)$\min\{a(1-b),b(1-c),c(1-a)\} \le \sqrt[3]{a(1-b)b(1-c)c(1-a)} = \sqrt[3]{a(1-a)b(1-b)c(1-c )}$ $ \le \sqrt[3]{(\frac{1}{2})^2(\frac{1}{2})^2(\frac{1}{2})^2} = \frac{1}{4}$ 2)$L=\max\{a(1-b),b(1-c),c(1-a)\} \ge \frac{a(1-b)+b(1-c)+c(1-a)}{3} = \frac{1-(ab+bc+ca)}{3}$ By Cauchy Schwartz, $(a+b+c)^2 \ge 3(ab+bc+ca) \implies ab+bc+ca \le \frac{1}{3}$ So $L \ge \frac{1-\frac{1}{3}}{3} = \frac{2}{9}$

In order to prove that $\min\{a(1-b),b(1-c),c(1-a)\}\leq \frac{1}{4}$, it suffices to prove that $a(1-b)b(1-c)c(1-a)\leq \frac 1{64}$. We have $a(1-a)\leq \left(\frac{a+(1-a)}{2}\right)^2=\frac 14$, so $a(1-b)b(1-c)c(1-a)\leq \frac 1{64}$. In order to prove that $\max\{a(1-b),b(1-c),c(1-a)\}\geq \frac{2}{9}$, it suffices to prove that $a(1-b)+b(1-c)+c(1-a)\ge \frac 23\Leftrightarrow a+b+c-ab-bc-ca\ge \frac 23\Leftrightarrow ab+bc+ca\leq \frac 13$. We have $ab+bc+ca\leq \frac{(a+b+c)^2}{3}=\frac 13$, as desired.

Let $a,b,c$ be three real numbers of $[0,1]$. Prove that $$min\left\{a(1-b),b(1-c),c(1-a)\right\}\leq\frac{1}{4}$$

Let $a, b, c$ positive reals such that $a+b+c=1$. Prove that$$\min\{a(1-b),b(1-c),c(1-a)\}\leq \frac{2}{9}\leq \max\{a(1-b),b(1-c),c(1-a)\}$$