Final Round of National Olympiad

If some two numbers among $x_1$, $x_2$, $x_3$, $x_4$ are equal then $K=0$ which is not maximal. Thus assume w.l.o.g. that $x_1>x_2>x_3>x_4$. Applying AM-GM for $x_1-x_2$, $x_2-x_3$ and $x_3-x_4$ gives $$\sqrt[3]{(x_1-x_2)(x_2-x_3)(x_3-x_4)}\le \frac{(x_1-x_2)+(x_2-x_3)+(x_3-x_4)}{3}=\frac{x_1-x_4}{3}\le \frac{1}{3},$$ i.e., $|x_1-x_2|\cdot |x_2-x_3|\cdot |x_3-x_4|\le \frac{1}{27}$. Among the remaining factors $|x_1-x_3|$, $|x_1-x_4|$, $|x_2-x_4|$, at least one is less than 1. Hence we conclude the left-hand inequality needed.

For the second inequality, note that if $x_1=1$, $x_2=\frac{3}{4}$, $x_3=\frac{1}{4}$, $x_4=0$ then $$K=\frac{1}{4}\cdot \frac{3}{4}\cdot 1\cdot \frac{1}{2}\cdot \frac{3}{4}\cdot \frac{1}{4}=\frac{9}{512}>\frac{4}{243},$$ since $9\cdot 243=2187>2048=4\cdot 512$.

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Alternative solution.

W.l.o.g., assume the inequalities $x_1>x_2>x_3>x_4$. In addition, assume that $x_1=1$ and $x_4=0$ as otherwise $K$ can be made larger. Substituting $x_2=y$ and $x_3=z$ for simplicity, one obtains $$K=(1-y)(1-z)(y-z)yz=(y-z)\cdot (1-y)z\cdot (1-z)y.$$ Consider pairs $(y,z)$ with $y-z$ fixed. The sums $(1-y)+z=1-(y-z)$ and $(1-z)+y=1+(y-z)$ are then also fixed. The product of two numbers with fixed sum is the largest if the numbers are equal; thus the product $(1-y)z$ is the largest in the case $y+z=1$ and the product $(1-z)y$ is the largest in the same case $y+z=1$. Consequently, also $K$ obtains its largest value in the case $y+z=1$. Substituting $1-z$ at place of $y$, one gets $K=z^2(1-z{)}^2(1-2z)=(z(1-z){)}^2(1-2z)$. Let $f(z)=(z(1-z){)}^2(1-2z)$, then $f^{\prime }(z)=2z(1-z)(1-5z+5z^2)$. The roots of $f^{\prime }$ within $(0,1)$ are the roots of the quadratic polynomial $5z^2-5z+1$, namely $\frac{5-\sqrt{5}}{10}$ and $\frac{5+\sqrt{5}}{10}=1-\frac{5-\sqrt{5}}{10}$. As $f(0)=f(\frac{1}{2})=0$ and $f(z)>0$ whenever $0<z<\frac{1}{2}$, the maximum of $f$ is achieved at $z=\frac{5-\sqrt{5}}{10}$. Thus the maximum value of $K$ is $f(\frac{5-\sqrt{5}}{10})=\frac{\sqrt{5}}{125}$. This number $\frac{\sqrt{5}}{125}$ satisfies both inequalities of the problem.