Can we prove the inequality without opening the parentheses? $(x+y+z)(xy+yz+xz)(x^2+y^2+z^2)≥6(x^2+y^2+z^2)+3(xy+yz+xz)$

Question

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

$$(x+y+z)(xy+yz+xz)(x^2+y^2+z^2)≥6(x^2+y^2+z^2)+3(xy+yz+xz)$$

I tried to use the following inequalities:

$$x^2+y^2+z^2≥xy+yz+xz$$

and The Cauchy–Schwarz inequality:

$$x^2+y^2+z^2≥\frac{(x+y+z)^2}{3}$$

$$x^2+y^2+z^2≥x+y+z$$

I also tried

$$x+y+z ≥3\\ x^2+y^2+z^2≥3\\ xy+yz+xz≥3$$

But, I couldn't make progress.

I want to solve this inequality without expansion. That is, without opening the parentheses. Is this possible?

Answer 1

Increase the right hand side to $9(x^2+y^2+z^2)$ by switching $xy+yz+zx$ to $x^2+y^2+z^2$ (Cauchy).

Divide the left hand side by $xyz=1$.

This leads to $(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq 9$, which is easy to show. Namely, $x+y+z\geq 3\sqrt[3]{xyz}$ by AM-GM and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq 3/\sqrt[3]{xyz}$ by HM-GM. This also shows that equality holds iff $x=y=z=1$.

Answer 2

Rewrite the left hand side as follows: $$ (x+y+z)(xy+yz+xz)(x^2+y^2+z^2)= \\ =\left(\frac{2}{3}(x+y+z)(xy+yz+zx)\right)\cdot(x^2+y^2+z^2)+ \\ +\left(\frac{1}{3}(x+y+z)(x^2+y^2+z^2)\right)\cdot(xy+yz+zx). $$ Can you continue now?