Show that $ \max\big\{ (1-x)(1-y) , xy , x(1-y) + y(1-x)\big \} \geq \dfrac49$ for $x,y \in [0,1]$.

Let $s:=x+y$ and $m:=xy$. We have $$0\leq s\leq 2\text{ and }0\leq m\leq \frac{s^2}{4}\,.$$ The three functions to be compared are $$a:=(1-x)(1-y)=1-s+m\,,\,\,b:=xy=m\,,\text{ and }c:=x(1-y)+y(1-x)=s-2m\,.$$ If $0\leq s\leq \dfrac{2}{3}$, then $m\leq \dfrac{s^2}{4}\leq \dfrac{1}{9}$, implying $$a+c=1-m\geq \frac{8}{9}\,,\text{ whence }a\geq \frac{4}{9}\text{ or }c\geq \frac{4}{9}\,.$$ If $\dfrac{2}{3}\leq s\leq \dfrac{4}{3}$, then $|s-1|\leq \dfrac13$, whence $$(s-1)^2\leq \frac{1}{9}\,,\text{ or }s-\frac{s^2}{2}\geq \frac{4}{9}\,.$$ Ergo, $$c=s-2m\geq s-\frac{s^2}{2} \geq \frac{4}{9}\,.$$ Finally, suppose that $\dfrac{4}{3}\leq s\leq 2$. Then, $$2b+c=s\geq \frac{4}{3}\,,\text{ making }b\geq \frac{4}{9}\text{ or }c\geq \frac{4}{9}\,.$$ Hence, we do have $$\frac49\leq \max\big\{(1-x)(1-y),xy,x(1-y)+y(1-x)\big\}\leq 1\text{ for all }x,y\in[0,1]\,,$$ whose left-hand side is an equality is an equality iff $(x,y)=\left(\dfrac13,\dfrac13\right)$ or $(x,y)=\left(\dfrac23,\dfrac23\right)$, and whose right-hand side is an equality iff $x\in\{0,1\}$ and $y\in\{0,1\}$.

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Interestingly, we also have $$0\leq \min\big\{(1-x)(1-y),xy,x(1-y)+y(1-x)\big\}\leq \frac14\text{ for all }x,y\in[0,1]\,,$$ where the left-hand side is an equality iff $x\in\{0,1\}$ or $y\in\{0,1\}$, and the right-hand side is an equality iff $(x,y)=\left(\dfrac12,\dfrac12\right)$. Here is a proof of the right-hand side inequality above. If $0\leq s\leq 1$, then we get $$b=m\leq \frac{s^2}{4}\leq \dfrac14\,.$$ If $1\leq s\leq 2$, then we have $$a=1-s+m\leq 1-s+\frac{s^2}{4}=\frac{(2-s)^2}{4}\leq \frac{1}{4}\,.$$

Furthermore, the median satisfies $$0\leq\text{med}\big((1-x)(1-y),xy,x(1-y)+y(1-x)\big)\leq \frac12\,.$$ The left-hand side inequality becomes an equality iff $x\in\{0,1\}$ and $y\in\{0,1\}$. The right-hand side inequality becomes an equality iff $\{x,y\}=\left\{0,\dfrac12\right\}$ or $\{x,y\}=\left\{\dfrac12,1\right\}$.

Idea:

Instead of $y$ write $a$. Now we have $3$ linear functions $$f(x) = x(1-2a)+a$$ $$g(x) =x(a-1)+1-a$$ $$h(x) = ax$$

Now if $a\geq {2\over 3}$ then $f$ is entirely above $g$ but $f$ and $h$ intersect at $x_0={a\over 3a-1}$

In this case $$f(x_0) = {a^2\over 3a-1}\geq {4\over 9}$$

The same porcedure we do for $a\leq 1/3$ and $1/3\leq a\leq 2/3$...