MMO2025 Round 4

Without loss of generality we may assume that $a,b,c,d\ge 0$. By dividing $2$, it is enough to prove the inequality for the numbers $0\le a,b,c,d\le 1$: $$\frac{a^3}{b^2+1}+\frac{b^3}{c^2+1}+\frac{c^3}{d^2+1}+\frac{d^3}{a^2+1}\le 2.$$ Now, suppose that $a$ is the largest among the $a,b,c,d$. Then, $$\frac{a^3}{b^2+1}+\frac{b^3}{c^2+1}\le \frac{1}{2}+\frac{a^3}{c^2+1}.$$ Indeed, clearing denominators and simplifying gives $$2a^3{b}^2+b^2{c}^2+b^2+c^2+1\ge 2a^3{c}^2+2b^5+2b^3$$ This follows from the inequality $b^2{c}^2+1\ge b^2+c^2$. $b^2(a^3-b^3)+b^2(1-b)+c^2(1-a^3)\ge 0$ Similarly, $$\frac{a^3}{c^2+1}+\frac{c^3}{d^2+1}\le \frac{1}{2}+\frac{a^3}{d^2+1}\text{and}\frac{a^3}{d^2+1}+\frac{d^3}{a^2+1}\le \frac{1}{2}+\frac{a^3}{a^2+1}.$$ Clearly, $2a^3\le a^2+1$, so the problem is solved.