Prove the inequality $4+xy+yz+zx \ge 7xyz$

Question

Let $x,y,z$ be non negative real numbers such that $x+y+z=3$, prove the following inequality:

$$4+xy+yz+zx \ge 7xyz$$

I tried MV and taking out one variable but I got nothing

Answer 1

Hint Use AM-GM to show $$1\ge xyz, \quad (x+y+z)(xy+yz+zx)\ge9xyz$$

Answer 2

Alternative approach:

We have that $1=\frac{x+y+z}{3}\geq\sqrt[3]{xyz}$, therefore $xyz\leq 1$.

Since $xyz\leq 1$, we have $4 \geq 4xyz$.

Also, by AM-HM, we have:

$$1=\frac{x+y+z}{3} \geq \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$$

Therefore

$$\frac{1}{3} \geq \frac{1}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$$ $$3 \leq \frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$

$$3xyz \leq yz+xz+xy$$

Adding the latter inequality and $4 \geq 4xyz$ gives the desired inequality.

Answer 3

Note that because $x+y+z=3$ we have $1\geq xyz$ and $4\geq 4xyz$.

In addition, $xy+yz+xz \geq 3(xyz)^\frac{2}{3}$ but because $1\geq xyz$ we have $1\geq xyz^\frac{1}{3}$ and $3(xyz)^\frac{2}{3}\geq3xyz$. Hence, $xy+yz+xz \geq 3xyz$.

Adding the two inequalities we have $4+xy+yz+xz \geq 7xyz$.