Bump.....

jasperE3 wrote: Let $m$ be the smallest of the four numbers $1,x^9,y^9,z^7$, where $x,y,z$ are nonnegative numbers. Prove that $m\le xy^2z^3$. $$xy^2z^3\geq m^{\frac{1}{9}}m^{\frac{2}{9}}m^{\frac{3}{7}}=m^{\frac{16}{21}}\geq m.$$

Very nice