Given three non-negative numbers $x, y, z$ so that $x+ y+ z= 100$ and $x, y, z\notin\left ( 33, 34 \right )$, show that $xyz\leq 33\cdot 34\cdot 33$

Question

Given three non-negative numbers $x, y, z$ so that $x+ y+ z= 100$ and $x, y, z\notin\left ( 33, 34 \right )$, show that $$xyz\leq 33\cdot 34\cdot 33$$ I would like to see a solution that is different from the following approach by a Vietnamese student:
Existing solution:
Without loss of generality, assume $x\geq y\geq z$. Then since none of $x, y, z$ lies in the open interval $\left ( 33, 34 \right )$, it follows that $x\geq 34$ and $z\leq 33$. Applying the AM-GM inequality: $$\frac{x}{34}\frac{y}{33}\frac{z}{33}\leq\left ( \frac{\frac{x}{34}+ \frac{y}{33}+ \frac{z}{33}}{3} \right )^{3}= \left ( \frac{34(x+ y+ z)- x}{3\cdot 34\cdot 33} \right )^{3}\leq 1$$ Thus $xyz\leq 33\cdot 34\cdot 33$.

Can you provide an alternative proof that avoids the same technique or structure—perhaps using different inequalities, optimization strategies, or case analysis? Thank you!

Answer 1

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!