Proving $\frac{x^2+y^2+z^2}{x+y+z}+\frac 32\sqrt[3]\frac{xy+yz+xz}{x+y+z}\geq \frac 52$, for positive values with $xyz=1$, without expansion?

Question

Let, $x,y,z>0$ and $xyz=1$, then prove that

$$\frac{x^2+y^2+z^2}{x+y+z}+\frac 32\sqrt[3]\frac{xy+yz+xz}{x+y+z}\geq \frac 52$$

I know that $$x^2+y^2+z^2\geq x+y+z$$ by Cauchy-Schwarz. So, $$\frac{x^2+y^2+z^2}{x+y+z}\geq 1$$

But, $xy+yz+xz\geq x+y+z$ is not always true. My question is, is there a way to prove this inequality without expansion? I mean, I want to avoid getting the cube of each side. Is it possible?

Answer 1

Here is a solution using only various mean inequalities:

First use AM-GM:

$$2\frac{x^2+y^2+z^2}{x+y+z}+3\sqrt[3]{\frac{xy+yz+zx}{x+y+z}}\ge5\sqrt[5]{\frac{(x^2+y^2+z^2)^2(xy+yz+zx)}{(x+y+z)^3}}$$

Then using $xyz=1$:

$$\frac{(x^2+y^2+z^2)^2(xy+yz+zx)}{(x+y+z)^3}=\frac{(x^2+y^2+z^2)^2}{(x+y+z)^4}(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$$

Using AM-HM:

$$(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge 9$$

Using AM-QM:

$$\frac{(x^2+y^2+z^2)^2}{(x+y+z)^4}\ge\frac{1}{9}$$

Equality takes place for $x=y=z=1$.