<s>[INST] 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


 [/INST] 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 $$4v^5 +94u^2 v^3 -7uv^4 \ge 2\sqrt{(4v^5).(94u^2 v^3)} -7uv^4 =(4\sqrt{94}-7)uv^4 >0$$ 
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