Find the smallest $C>0$ such that $xy^2z^3 \leq C(x+y+z)^6$ when $x,y,z \geq 0$

Question

The author recommended that I first find the smallest $C>0$ while also having $x+y+z=c$ where $c>0$. I'm not really sure how this is going to help me.

How should I go about solving this problem? I'm not looking for someone to solve it, I just want some help getting started.

Answer 1

Using AM-GM inequality: $ x+y +z = x + y/2 +y/2 + z/3+z/3+z/3 \ge 6\sqrt[6]{xy^2/4z^3/27} = 6\sqrt[6]{xy^2z^3/108} \implies C = 4 \cdot 27 /6^6 = 1/432$

Answer 2

So in general for $x_i\ge 0$, $\lambda_i>0$ we have,denoting by $\Lambda = \sum_k \lambda_k$ $$\sum x_i = \sum_i\frac{\lambda_i}{\Lambda}\cdot (\frac{\Lambda}{\lambda_i} x_i) \ge \prod_i (\frac{\Lambda}{\lambda_i} x_i)^{\frac{\lambda_i}{\Lambda}}$$ and so $$\frac{\prod \lambda_i^{\lambda_i}}{\Lambda^{\Lambda}}\cdot(\sum x_i)^{\Lambda}\ge \prod_i x_i^{\lambda_i}$$ and the constant $C_{(\lambda)}=\frac{\prod \lambda_i^{\lambda_i}}{\Lambda^{\Lambda}}$ is the best possible.

In our case $$\frac{1^1\cdot 2^2 \cdot 3^3 }{6^6}\cdot (x+y+z)^6 \ge xy^2 z^3$$ and $\frac{1^1\cdot 2^2 \cdot 3^3 }{6^6}=\frac{1}{432}$.

Looking at the author's recommendation, it is maximizing the product $x y^2 z^3$ knowing that $x+y+z=c$ is constant, using Lagrange multipliers. At the point of maximm get $$\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$$ and so $=\frac{x+y+z}{6}$, hence the inequality.