Given three non-negative numbers $x, y, z$ so that $x+ y+ z= 100$ and $x, y, z\notin (\!33, 34\!)$. Prove that
$$xyz\leqq 33\cdot 34\cdot 33$$
I want to see another solution that is not as same as the following one due to one Vietnamese student
Wlog $x\geqq y\geqq z$ then $x\geqq 34$ and $z\leqq 33$, from am-gm inequality
$$\frac{x}{34}\frac{y}{33}\frac{z}{33}\leqq \left ( \frac{\frac{x}{34}+ \frac{y}{33}+ \frac{z}{33}}{3} \right )^{3}= \left ( \frac{34(x+ y+ z)- x}{3\cdot 34\cdot 33} \right )^{3}\leqq 1$$
so $xyz\leqq 33\cdot 34\cdot 33$. I took the source from DOTOANNANG(@HaiDangel). I hope you like it, thanks
Best Answer
Let $x\geq y\geq z$.
Thus, $x\geq34$ and $z\leq33$ and since $$x+y=100-z\geq100-33=34+33,$$ we obtain $$(x,y,z)\succ(34,33,33)$$
and since $f(x)=e^x$ increases and $\ln$ is a concave function, by Karamata we obtain: $$xyz=e^{\ln{x}+\ln{y}+\ln{z}}\leq e^{\ln34+\ln33+\ln33}=34\cdot33^2$$ and we are done!