parmenides51 wrote: Find all the real numbers $k$ that have the following property: For any non-zero real numbers $a$ and $b$, it is true that at least one of the following numbers: $$a, b,\frac{5}{a^2}+\frac{6}{b^3}$$is less than or equal to $k$. From $a = b = 5/a^2 + 6/b^3$, we have $a=b=2$ or $a=b=-1$. Let us prove that $\min(a, b, 5/a^2 + 6/b^3) \le 2$ for all real numbers $a, b$ with $ab\ne 0$. If $a, b > 2$, we have $5/a^2 + 6/b^3 \le 5/2^2 + 6/2^3 = 2$.