If $x,y,z>0.$Prove: $(x+y+z) \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right) \geq9\sqrt[]\frac{x^2+y^2+z^2}{xy+yz+zx}$ If $x,y,z>0.$Prove:
$$(x+y+z) \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)\geq9\sqrt[]\frac{x^2+y^2+z^2}{xy+yz+zx}$$
I was not able to solve this problem instead I could solve similar inequality when we have two variable.I assumed $y=tx$ and uesd derivative.
Can this be generalized as:

If ${a_i>0}\quad(i=1,2,...,n)$
$$\sum_{i=1}^n a_{i} \sum_{i=1}^n \frac{1}{a_{i}}\geq n^2\sqrt[]\frac{\sum_{i=1}^n a^2_{i} }{\sum_{i=1}^n a_{i}a_{i+1} }$$
$a_{n+1}=a_{1}$

Question from Jalil Hajimir



Here I give a proof by using the standard pqr method.
Proof: Let $p = x+y+z$, $q = xy+yz+zx$ and $r = xyz$.
We will use the following facts (see [1], Facts N12 and N6):
(i) $q^3 + 9r^2 \ge 4pqr$.
(ii) $q^3 \ge 27r^2$.
We need to prove that
$$\frac{pq}{r} \ge 9 \sqrt{\frac{p^2-2q}{q}}$$
or
$$\frac{p^2q^2}{r^2} \ge 81 \frac{p^2-2q}{q}$$
or
$$162qr^2 - (81r^2 - q^3)p^2 \ge 0.$$
There are two possible cases:
1) If $81r^2 - q^3 > 0$:  From Fact (i), we have $\frac{q^3+9r^2}{4qr} \ge p$. It suffices to prove that
$$162qr^2 - (81r^2 - q^3)\left(\frac{q^3+9r^2}{4qr}\right)^2 \ge 0$$
or
$$\frac{(q^3 - 9r^2)(q^3 - 27r^2)^2}{16q^2r^2} \ge 0.$$
It is true by using Fact (ii).
2) If $81r^2 - q^3 \le 0$, clearly the inequality is true.
We are done.
Reference:
[1] Zdravko Cvetkovski, "Inequalities Theorems, Techniques and Selected Problems", Ch. 14.
https://keoserey.files.wordpress.com/2012/07/zdravko-cvetkovski-inequalities-theorems_-techniques-and-selected-problems.pdf
Remark: In Cvetkovski's book, chapter 14, Page 138, Cvetkovski gave Facts N1 through N13. There is a typo:
Fact N10 should be $2q^3 + 9r^2 \ge 7pqr$ (rather than $2p^3 + 9r^2 \ge 7pqr$ which is not true for $a=4, b=3, c=2$).
