Iranian Mathematical Olympiad

Putting $C=\frac{9}{1000}$, $n=1400$. Assume to the contrary that for all $z\ge y\ge x$ we have $$(z-y)(y-x)(z-x)\ge C(1+x^4+y^4+z^4).$$ Notice that $4(y-x)(z-y)\le (y-x+z-y{)}^2=(z-x{)}^2$. Whence, $$(z-x{)}^3\ge 4(z-y)(y-x)(z-x)\ge 4C(1+x^4+y^4+z^4)\ge 4C(1+x^4+z^4)\ge 4C.$$ It follows that $z-x\ge \sqrt[3]{4C}$. On the other hand, we can also deduce that $$\frac{(z-x{)}^3}{x^4+z^4}\ge 4C.$$ Since $2(x^4+z^4)\ge (x^2+z^2{)}^2$ and $2(x^2+z^2)\ge (x-z{)}^2$. Therefore, $8(x^4+z^4)\ge (x-z{)}^4$. We would obtain $$\frac{8}{z-x}=\frac{8(z-x{)}^3}{(x-z{)}^4}\ge \frac{(z-x{)}^3}{x^4+z^4}\ge 4C.$$ That is, $$\sqrt[3]{4C}\le z-x<\frac{2}{C}.$$ Finally, assume that $x_1\le \cdots \le x_n$ are given. We would then have $x_{k+2}-x_k>\sqrt[3]{4C}$. Hence, $$\frac{2}{C}>x_n-x_1\ge \lfloor \frac{n-1}{2}\rfloor \sqrt[3]{4C}.$$ Hence, $$C<\frac{\sqrt[3]{2}}{{\lfloor \frac{n-1}{2}\rfloor }^{\frac{3}{2}}}.$$ Yielding, $C<\frac{87}{10000}$. A contradiction. Thus, we are done. ■