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


 A: 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 $a=4, b=3, c=2$).
A: Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, we need to prove that $f(w^3)\geq0,$ where
$$f(w^3)=\frac{uv^2}{w^3}-\sqrt{\frac{3u^2-2v^2}{v^2}}.$$
We see that $f$ decreases, which says that it's enough to prove our inequality for a maximal value of $w^3$, which by $uvw$ ( https://artofproblemsolving.com/community/c6h278791 )
happens for equality case of two variables.
Since our inequality is homogeneous, we can assume $y=z=1,$ which gives
$$(x+2)^2(2x+1)^3\geq81x^2(x^2+2)$$ or
$$(x-1)^2(8x^3-21x^2+36x+4)\geq0,$$ which is obvious.
A: I have a solution using Buffalo way, but it's ugly! I'm sorry about that!
Solution:
Without loss of generality, assume that $x=\min\{x,y,z\}$.
Let $x=a$, $y=a+u$, $z=a+v$ so $a>0$; $u,v \geq 0$
We need to prove: $$(x+y+z)^2 (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2 - 81\frac{(x^2 +y^2 +z^2)}{xy+yz+zx} \geq 0$$
After reduction of many fractions to a common denominator, we need to prove:
$$27a^6(u^2 -uv+v^2)+18a^5 (u+v)^3 +3a^4 (u^4 +13u^3 v+78u^2 v^2+13uv^3 +v^4 )+2a^3(4u^5 -7u^4 v+94u^3 v^2 +94u^2 v^3 -7uv^4 +4v^5)+3a^2 uv(4u^4 +4u^3 v+57u^2 v^2 +4uv^3 +4v^4)+6au^2 v^2(u^3 +4uv(u+v)+v^3)+u^3 v^3 (u+v)^2 \geq 0$$
Because: $u^2 -uv+v^2 \geq 0$; $(u+v)^3 \geq 0$; $u^4 +13u^3 v+78u^2 v^2+13uv^3 +v^4 \geq 0$, $uv(4u^4 +4u^3 v+57u^2 v^2 +4uv^3 +4v^4)$;$(u^3 +4uv(u+v)+v^3)\geq 0$; $u^3 v^3 (u+v)^2 \geq 0$ 
So it suffices to prove: $4u^5 -7u^4 v+94u^3 v^2 +94u^2 v^3 -7uv^4 +4v^5 \geq 0$
But it's obvious true by AM-GM: $$4u^5+94u^3 v^2 -7u^4 v \geq 2\sqrt{(4u^5).(94u^3 v^2)} - 7u^4 v =(4\sqrt{94}-7)u^4 v >0$$
And $$4v^5 +94u^2 v^3 -7uv^4 \ge 2\sqrt{(4v^5).(94u^2 v^3)} -7uv^4 =(4\sqrt{94}-7)uv^4 >0$$ 
