Yes but it's not mine, i need new method :)

Let x1=a2a1;x2=a3a2;...;xn=a1an

We need prove a1a1+a2+a3+a2a2+a3+a4+...+anan+a1+a2>1

It's right because n>3

 and ai+ai+1+ai+2<a1+a2+...+an∀i

Yes but it's not mine, i need new method :)

Let \(x_1=\dfrac{a_2}{a_1};x_2=\dfrac{a_3}{a_2};...;x_n=\dfrac{a_1}{a_n}\)

We need prove \(\dfrac{a_1}{a_1+a_2+a_3}+\dfrac{a_2}{a_2+a_3+a_4}+...+\dfrac{a_n}{a_n+a_1+a_2}>1\)

It's right because \(n>3\) and \(a_i+a_{i+1}+a_{i+2}< a_1+a_2+...+a_n\forall i\)

Ace Legona : you can share a answer of problem.

This is a good and difficult problem.