show this inequality $a_{1}+a_{2}+\cdots+a_{n}\ge\sqrt{3}(a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n}a_{1})$ 
let $a_{1},a_{2},\cdots,a_{n}\ge 0,n\ge 3$,and such $$a^2_{1}+a^2_{2}+\cdots+a^2_{n}=1$$
  show that
  $$a_{1}+a_{2}+\cdots+a_{n}\ge\sqrt{3}(a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n}a_{1})$$

I can prove when $n=3$, it need to prove
$$a_{1}+a_{2}+a_{3}\ge\sqrt{3}(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1})$$
where $a^2_{1}+a^2_{2}+a^2_{3}=1$.
since $$ (a_{1}+a_{2}+a_{3})^2\ge 3(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1})\tag{1}$$
and 
$$1=a^2_{1}+a^2_{2}+a^2_{3}\ge a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}\tag{2}$$
$(1)\times (2)$ we have
$$(a_{1}+a_{2}+a_{3})^2\ge 3(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1})^2$$
so we have$$a_{1}+a_{2}+a_{3}\ge\sqrt{3}(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1})$$
But for $n\ge 4$ ,I want to show 
$$f(a_{1},a_{2},\cdots,a_{n})-f(a_{1},a_{2},a_{3},\cdots,a_{n-2},\sqrt{a^2_{n-1}+a^2_{n}},0)\ge 0$$,where
$$a_{1}=\max(a_{1},a_{2},\cdots,a_{n}).f(a_{1},a_{2},\cdots,a_{n})=\dfrac{1}{\sqrt{3}}(a_{1}+\cdots+a_{n})-(a_{1}a_{2}+\cdots+a_{n}a_{1})$$
and
$$f(a_{1},a_{2},\cdots,a_{n})-f(a_{1},a_{2},a_{3},\cdots,a_{n-2},\sqrt{a^2_{n-1}+a^2_{n}},0)$$$$=\dfrac{1}{\sqrt{3}}(a_{n-1}+a_{n}-\sqrt{a^2_{n-1}+a^2_{n}})+a_{n-2}\sqrt{a^2_{n}+a^2_{n-1}}-a_{n-2}a_{n-1}-a_{n-1}a_{n}-a_{n}a_{1}$$I can't prove it
 A: Observe that $f(a_1,a_2,...,a_n)=a^2_{1}+a^2_{2}+\cdots+a^2_{n}$ is a symmetric polynomial,  so you may write it in terms of the elementary symmetric polynomials.
As a matter of fact, you may write $a^2_{1}+a^2_{2}+\cdots+a^2_{n}=(a_1+a_2+...+a_n)^2-2(a_1a_2+a_1a_3+...+a_{n-1}a_n)$
Therefore we have:
$(a_1+a_2+...+a_n)^2-2(a_1a_2+a_1a_3+...+a_{n-1}a_n)=1$, which you may rearrange to get:
$(a_1+a_2+...+a_n)^2 = 1+2(a_1a_2+a_1a_3+...+a_{n-1}a_n)$.
You may now use the rearrangement inequality : (look at this : https://brilliant.org/wiki/rearrangement-inequality/) to conclude that $1=a^2_{1}+a^2_{2}+\cdots+a^2_{n} \geq a_1a_2+a_2a_3+...+a_{n-1}a_{n}$
Putting everything together we get: $(a_1+a_2+...+a_n)^2 = 1+2(a_1a_2+a_1a_3+...+a_{n-1}a_n) $
$\geq (a_1a_2+a_2a_3+...+a_{n-1}a_{n})+2(a_1a_2+a_1a_3+...a_{n-1}a_{n} )$
$\geq(a_1a_2+a_2a_3+...+a_{n-1}a_{n})+2(a_1a_2+a_2a_3+...+a_{n-1}a_{n}) $
=$3(a_1a_2+a_2a_3+...+a_{n-1}a_{n}) $
(on a sidenote, I'm not 100% sure what $(a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n}a_{1})$ means in your question, as in I'm not sure whether this is $\sum\prod_{1 \leq i <j \leq n} a_ia_j$, or simply $a_1a_2+a_2a_3+a_3a_4+...a_{n-1}a_{n}$ )
Edit by cr001:
$$a_1a_2+a_2a_3+...+a_na_1\leq {a_1}^2+{a_2}^2+...+{a_n}^2\leq 1\implies$$
$$a_1a_2+a_2a_3+...+a_na_1 \geq (a_1a_2+a_2a_3+...+a_na_1)^2$$
which concludes the proof.
A: For $n=3$ you have a proof.
We'll prove that for all $n\geq4$ the following stronger inequality is true:
$$\sum_{k=1}^na_k\geq2\sum_{k=1}^na_ka_{k+1},$$ where $a_{n+1}=a_1$.
Indeed, we need to prove that
$$\sum_{k=1}^na_k^2\left(\sum_{k=1}^na_k\right)^2\geq4\left(\sum_{k=1}^na_ka_{k+1}\right)^2,$$ which is true because
$$\sum_{k=1}^na_k^2-\sum_{k=1}^na_ka_{k+1}=\frac{1}{2}\sum_{k=1}^n\left(a_k-a_{k+1}\right)^2\geq0$$ and
$$\left(\sum_{k=1}^na_k\right)^2\geq4\sum_{k=1}^na_ka_{k+1}$$  is true by AM-GM:
$$\sum_{k=1}^na_ka_{k+1}\leq(a_1+a_3+...)(a_2+a_4+...)\leq\left(\frac{\sum\limits_{k=1}^na_k}{2}\right)^2$$ because for odd $n$ we can assume that $a_1=\min\limits_{i}\{a_i\}$.
For example, for $n=5$ it works so:
$$a_1a_2+a_2a_3+a_3a_4+a_4a_5+a_5a_1\leq $$
$$\leq a_1a_2+a_2a_3+a_3a_4+a_4a_5+a_5a_2+a_4a_1=$$
$$=(a_1+a_3+a_5)(a_2+a_4)\leq\left(\frac{a_1+a_3+a_5+a_2+a_4}{2}\right)^2.$$
