IMO Team Selection Test 1

We will show that the minimum value is $\frac{4}{9}$. This value can be attained by taking $x=y=\frac{2}{3}$. Then we have $xy=\frac{4}{9}$, $(x-1)(y-1)=\frac{1}{9}$, and $x+y-2xy=\frac{4}{9}$, and the maximum is indeed $\frac{4}{9}$.

Now we will prove that $M(x,y)\ge \frac{4}{9}$ for all $x$ and $y$. Let $a=xy$, $b=(x-1)(y-1)$, and $c=x+y-2xy$. If we replace $x$ and $y$ by $1-x$ and $1-y$, then $a$ and $b$ will be interchanged and $c$ stays the same, because $(1-x)+(1-y)-2(1-x)(1-y)=2-x-y-2+2x+2y-2xy=x+y-2xy$. Hence, $M(1-x,1-y)=M(x,y)$.

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 that we may assume without loss of generality that $x+y\ge 1$.

Now write $x+y=1+t$ with $t\ge 0$. We also have $t\le 1$, because $x,y\le 1$ and hence $x+y\le 2$. The inequality between the arithmetic and geometric mean yields $$xy\le {\left(\frac{x+y}{2}\right)}^2=\frac{(1+t{)}^2}{4}=\frac{t^2+2t+1}{4}.$$ We have $b=xy-x-y+1=xy-(1+t)+1=xy-t=a-t$, hence $b\le a$. Moreover, $$c=x+y-2xy\ge (1+t)-2\cdot \frac{t^2+2t+1}{4}=\frac{2+2t}{2}-\frac{t^2+2t+1}{2}=\frac{1-t^2}{2}.$$ If $t\le \frac{1}{3}$, then we have $c\ge \frac{1-t^2}{2}\ge \frac{1-\frac{1}{9}}{2}=\frac{4}{9}$ and hence $M(x,y)\ge \frac{4}{9}$ as well. The remaining case is $t>\frac{1}{3}$. We have $c=x+y-2xy=1+t-2a>\frac{4}{3}-2a$. Moreover, $M(x,y)\ge \max (a,\frac{4}{3}-2a)$, hence $$3M(x,y)\ge a+a+\left(\frac{4}{3}-2a\right)=\frac{4}{3},$$ which yields that $M(x,y)\ge \frac{4}{9}$.

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