How prove this Lower bound with $\sum_{i=1}^{n}\prod_{j=1}^{i}a_{j}$ 
Let $a_{i}\in [0,1]$,and $a_{1}\ge a_{2}\ge\cdots \ge a_{n}$,show that
  $$a_{1}+a_{1}a_{2}+a_{1}a_{2}a_{3}+\cdots+a_{1}a_{2}\cdots a_{n}\ge \sum_{i=1}^{n}\left(\dfrac{a_{1}+a_{2}+\cdots+a_{n}}{n}\right)^i$$

For $n=2$,we have $$a_{1}+a_{1}a_{2}\ge \dfrac{a_{1}+a_{2}}{2}+\left(\dfrac{a_{1}+a_{2}}{2}\right)^2$$
$$\Longleftrightarrow 2a_{1}+4a_{1}a_{2}-2a_{2}\ge (a_{1}+a_{2})^2$$
$$2\ge a_{1}-a_{2}$$it is clear 
I have try mathematical induction，Assume that $n=k$ is true,consider
$n=k+1$,then we have 
$$a_{1}+a_{1}a_{2}+\cdots+a_{1}a_{2}\cdots a_{n}+a_{1}a_{2}\cdots a_{n+1} \ge \sum_{i=1}^{n}\left(\dfrac{a_{1}+a_{2}+\cdots+a_{n}}{n}\right)^i+a_{1}a_{2}\cdots a_{n}a_{n+1}$$
so it must prove that
$$\sum_{i=1}^{n}\left(\dfrac{a_{1}+a_{2}+\cdots+a_{n}}{n}\right)^i+a_{1}a_{2}\cdots a_{n}a_{n+1}\ge \sum_{i=1}^{n+1}\left(\dfrac{a_{1}+a_{2}+\cdots+a_{n+1}}{n+1}\right)^{i}$$
then I can't 
so  How to prove it?
 A: More Elementary Proof (Except for the Use of Continuity)
Let $$g\left(a_1,a_2,\ldots,a_n\right):=\sum_{k=1}^n\,\prod_{j=1}^k\,a_j-\sum_{k=1}^n\,\left(\frac{1}{n}\,\sum_{j=1}^n\,a_j\right)^k$$ for all $a_1,a_2,\ldots,a_n\in[0,1]^n$ with $a_1\geq a_2\geq \ldots\geq a_n$.  We need to show that, for any $i\in\{1,2,\ldots,n-1\}$,
$$g\left(a_1,a_2,\ldots,a_n\right)\geq g\left(a_1,a_2,\ldots,a_{i-1},\frac{a_i+a_{i+1}}{2},\frac{a_i+a_{i+1}}{2},a_{i+2},\ldots,a_n\right)\,.$$
This is easy because if you subtract the left-hand side by the right-hand side of the inequality above, you get
$$\frac{a_i-a_{i+1}}{2}-\left(\frac{a_i-a_{i+1}}{2}\right)^2\,\left(1+a_{i+2}+a_{i+2}a_{i+3}+\ldots+a_{i+2}a_{i+3}\cdots a_n\right)\,.$$
The quantity above is greater than or equal to
$$\frac{a_i-a_{i+1}}{2}\left(1-\frac{1}{2}\left(1-a_{i+1}\right)\sum_{r=0}^{n-i-1}\,a_{i+1}^r\right)=\frac{a_i-a_{i+1}}{2}\left(\frac{1+a_{i+1}^{n-i}}{2}\right)\geq 0\,.$$
For any starting point $\textbf{a}^0\in[0,1]^n$, where $\textbf{a}^0:=\left(a_1^0,a_2^0,\ldots,a_n^0\right)$ satisfies $a_1^0\geq a_2^0\geq\ldots\geq a_n^0$ with $M^0:=\max\limits_{i\in\{1,2,\ldots,n-1\}}\,\left(a^0_i-a^0_{i+1}\right)$, it can be easily seen that the pair-averaging procedure described in the paragraph above can be applied to get a sequence of points $\textbf{a}^1,\textbf{a}^2,\ldots$, in such a way that the value $M^l:=\max\limits_{i\in\{1,2,\ldots,n-1\}}\,\left(a^l_i-a^l_{i+1}\right)$, where $\textbf{a}^l:=\left(a_1^l,a_2^l,\ldots,a_n^l\right)$, satisfies
$$M^{l+(n-1)}\leq \frac{1}{2}\,M^l$$
for all $l=0,1,2,\ldots$.  That means $\textbf{a}^l$ tends to the limit $\bar{\textbf{a}}:=\left(\bar{a},\bar{a},\ldots,\bar{a}\right)$, with $\bar{a}:=\frac{1}{n}\,\sum_{j=1}^n\,a_j^0$, as $l$ grows to infinity.  By continuity of $g$, we have
$$g\left(\textbf{a}^0\right)\geq g\left(\bar{\textbf{a}}\right)=0\,.$$
Alternatively, suppose that an $n$-tuple $\left(b_1,b_2,\ldots,b_n\right)\in[0,1]^n$ with $b_1\geq b_2\geq \ldots \geq b_n$ is such that $g\left(b_1,b_2,\ldots,b_n\right)$ is minimized (observe, using the compactness argument and the continuity of $g$, that this tuple must exist).  By the pair-averaging argument, we conclude that $b_1=b_2=\ldots=b_n$, whence the minimum possible value of $g$ is $0$.
The above argument shows that the desired inequality is true.  Furthermore, the equality case is when $a_1=a_2=\ldots=a_n$.
A: I am not sure if there is a simpler answer to this inequality or no; but the solution I am going to present involves Schur convexity and majorization. 
Suppose that for each sequence $(a_1,\dots,a_n)\in[0,1]^n$, the sequence $(a_{[1]},\dots,a_{[n]})$ is the ordered version with $a_{[1]}\geq\dots\geq a_{[n]}$. Consider the following function:
$$
f(a_1,\dots,a_n)=a_{[1]}+a_{[1]}a_{[2]}+\dots+a_{[1]}a_{[2]}\dots a_{[n]}.
$$
On the other hand note that $(a_1,a_2,\dots,a_n)\succ (\overline a,\dots,\overline a)$ with $\overline a=\frac{a_1+\dots+a_n}{n}$ and $\succ$ is the majorization order.
It is enough to prove that the function $f$ is Schur convex. Because  $(a_1,a_2,\dots,a_n)\succ (\overline a,\dots,\overline a)$ implies:
$$
f(a_1,a_2,\dots,a_n)\geq f(\overline a,\dots,\overline a),
$$
which is another way of expressing the same inequality.

Proof of Schur convexity of $f$:
To prove the Schur convexity of $f$, we use Schur-Ostrowski criterion. It is enough to verify that:
$$
(a_i-a_j)(\frac{\partial f}{\partial a_i}-\frac{\partial f}{\partial a_j})\geq 0
$$
for all sequences in $[0,1]^n$. 
Without loss of generality we can assume $a_1\geq \dots \geq a_n$. Also it is enough to check it for $i=1$ and $j>1$ because the verification of other pairs boils down to this case. See
$$
\frac{\partial f}{\partial a_1}=1+a_2+a_2a_3+\dots+a_2a_3\dots a_n\\
\frac{\partial f}{\partial a_j}=\sum_{i=j}^n\frac{\prod_{k=1}^i a_k}{a_j}
$$
and (with the convention of $\prod_{k=i}^j a_k=1$ and $\sum_{k=i}^j a_k=0$ for $j<i$):
$$
\frac{\partial f}{\partial a_1}-\frac{\partial f}{\partial a_j}=\sum_{i=1}^{j-1}{\prod_{k=2}^i a_k}+\sum_{i=j}^n(\prod_{k=1}^i a_k)(\frac{1}{a_1}-\frac{1}{a_j})\\
=\sum_{i=1}^{j-1}{\prod_{k=2}^i a_k}+\prod_{k=2}^{j-1} a_k(\sum_{i=j}^n\prod_{k=j+1}^i a_k)({a_j}-{a_1})\\
=\sum_{i=1}^{j-2}{\prod_{k=2}^i a_k}+\prod_{k=2}^{j-1}a_k\left(1- (\sum_{i=j}^n\prod_{k=j+1}^i a_k)({a_1}-{a_j})\right).
$$
See that the last expression is positive:
$$
({a_1}-{a_j})(\sum_{i=j}^n\prod_{k=j+1}^i a_k)\leq({1}-{a_{j+1}})(\sum_{i=j}^n\prod_{k=j+1}^i a_{j+1})=1-a_{j+1}^{n-j+1}\leq 1.
$$
This proves that $f$ satisfies Schur-Ostrowski criterion and implies the inequality.
