For real numbers $x$ and $y$, we define $M(x, y)$ as the maximum of the three numbers $x y, (x-1)(y-1)$, and $x+y-2 x y$. Determine the smallest possible value of $M(x, y)$ over all real numbers $x$ and $y$ with $0 \leq x, y \leq 1$.
We show that the minimum value is $\frac{4}{9}$. This value can be achieved by taking $x=y=\frac{2}{3}$. Then we have $x y=\frac{4}{9}, (x-1)(y-1)=\frac{1}{9}$, and $x+y-2 x y=\frac{4}{9}$, so the maximum is indeed $\frac{4}{9}$.

Now we will prove that $M(x, y) \geq \frac{4}{9}$ for all $x$ and $y$. Let $a=x y, b=(x-1)(y-1)$, and $c=x+y-2 x y$. If we replace $x$ and $y$ with $1-x$ and $1-y$, then $a$ and $b$ are swapped and $c$ remains the same, because $(1-x)+(1-y)-2(1-x)(1-y)=2-x-y-2+2 x+2 y-2 x y=x+y-2 x y$. Thus, $M(1-x, 1-y)=M(x, y)$. We have $x+y=2-(1-x)-(1-y)$, so at least one of $x+y$ and $(1-x)+(1-y)$ is greater than or equal to 1, which means we can assume without loss of generality that $x+y \geq 1$.

Now write $x+y=1+t$ with $t \geq 0$. We also have $t \leq 1$, because $x, y \leq 1$ so $x+y \leq 2$. From the inequality of the arithmetic and geometric means, we get

$$
x y \leq\left(\frac{x+y}{2}\right)^{2}=\frac{(1+t)^{2}}{4}=\frac{t^{2}+2 t+1}{4}
$$

We have $b=x y-x-y+1=x y-(1+t)+1=x y-t=a-t$, so $b \leq a$. Furthermore,

$$
c=x+y-2 x y \geq(1+t)-2 \cdot \frac{t^{2}+2 t+1}{4}=\frac{2+2 t}{2}-\frac{t^{2}+2 t+1}{2}=\frac{1-t^{2}}{2}.
$$

If $t \leq \frac{1}{3}$, then $c \geq \frac{1-t^{2}}{2} \geq \frac{1-\frac{1}{9}}{2}=\frac{4}{9}$, and thus $M(x, y) \geq \frac{4}{9}$.
The remaining case is $t>\frac{1}{3}$. We have $c=x+y-2 x y=1+t-2 a>\frac{4}{3}-2 a$. We have $M(x, y) \geq \max \left(a, \frac{4}{3}-2 a\right)$, so

$$
3 M(x, y) \geq a+a+\left(\frac{4}{3}-2 a\right)=\frac{4}{3},
$$

which implies that $M(x, y) \geq \frac{4}{9}$.
We conclude that the minimum value of $M(x, y)$ is $\frac{4}{9}$.