tnoodee wrote: Let $ x, y, z > 0 $ with $ x \leq 2, \;y \leq 3 \;$ and $ x+y+z = 11 $ Prove that $xyz \leq 36$ We have \[36-xyz=(y+9)(2-x)+6(3-y)+y(2-x)^2+x(3-y)^2\ge{0}\]

tnoodee wrote: Let $ x, y, z > 0 $ with $ x \leq 2, \;y \leq 3 \;$ and $ x+y+z = 11 $.Prove that $$xyz \leq 36$$ JBMO 2010 Shortlist A5

Very easy once you change x = 2 - f, y = 3 - m, z = 6 + m + f and then multiply the (2 - f)(3 - m)(6 + m + f). There is a small amount of work to reach the end, but this is still a nice and original solution. PS: Very nice problem, fun to solve, but definitely isn't an A5 problem.

$$xyz=\frac{3x.2y.z}{6} \le \frac{(3x+2y+z)^3}{162} =\frac{(2x+y+11)^3}{162} \le \frac{(2.2+3+11)^3}{162} =36$$

Not too hard. Just consider all of the ordered pairs that are possible. Or you could just do Ika789_Master's solution