If $a_a,a_2,\dots, a_n$ are in A.P. prove that $a_1a_2\dots a_n>(a_1a_n)^{n/2}$ If $a_a,a_2,\dots, a_n$ are positive reals in Arithmetic Progression, prove that $a_1a_2\dots a_n>(a_1a_n)^{n/2}$.  
$a_2-a_1=a_3-a_2=\dots=a_n-a_{n-1}=d$ say, then $a_n-a_1=(n-1)d$  
Is this approache correct? Some hint??
 A: Consider $$(a_1a_2\ldots a_n)^2>(a_1a_n)^n,$$ which is equivalent to $$(a_1a_n)^2(a_2a_{n-1})^2\ldots>(a_1a_n)^n.$$
So we must show that for all $i\in\{2,\ldots, \frac{n+1}{2}\}$ ,$$a_ia_{n-i+1}>a_1a_n.$$
$$a_ia_{n-i+1}=(a_1+(i-1)d)(a_1+(n-i)d)=a_1^2+a_1(i-1)d+a_1(n-i)d+(i-1)(n-i)d^2$$
We have $(i-1)(n-i)d^2=C>0$, which implies that
$$a_ia_{n-i+1}=a_1^2+a_1(n-1)d+C>a_1^2+a_1(n-1)d=a_1(a_1+(n-1)d)=a_1a_n$$, so since this inequality holds, then $(a_1a_2\ldots a_n)^2>(a_1a_n)^n$ holds and by taking the square root of both sides we get the desired result.
A: Let $$
X = \frac12( a_1 + a_n)
$$then for every $k$,
$$
X = \frac12( a_{1+k} + a_{n-k})
$$
Now the square of the left hand side is
\begin{align}
(a_1a_2\dots)^2 &= (a_1a_2\dots)(a_na_{n-1}\dots)
= a_1a_n\times a_2a_{n-1}\dots
\\&=
\left[X -  \frac12( a_1 - a_n)\right]\left[X +  \frac12( a_1 - a_n)\right]\times
\left[X -  \frac12( a_2 - a_{n-1})\right]\left[X +  \frac12( a_2 - a_{n-1})\right]
\dots
\\&= \left[X^2 -  \frac14( a_1 - a_n)^2\right]\times 
\left[X^2 -  \frac14( a_2 - a_{n-1})^2\right]\dots
\end{align}
where there are $2n$ factors in each equality until the last one, and $n$ if the last one.
For every $k$, $( a_{1+k} - a_{n-k})^2$ is maximin when $k =0$. So:
$$
(a_1a_2\dots)^2 \ge \left[X^2 -  \frac14( a_1 - a_n)^2\right]^n
$$which is your result.
A: Let $m=(a_1+a_n)/2$ and $p=(a_n-a_1)/2$. Take any $t$ with $0\le t\le p$ and notice that $(m-t)(m+t)=m^2-t^2\ge m^2-p^2=(m-p)(m+p)=a_1a_n$. 
Consider the case when $n$ is even, and list the arithmetic progression as 
$a_1,a_2,a_3,...,a_{k-2},a_{k-1},a_{k},a_{k+1},...,a_{n-2},a_{n-1},a_n$, where 
$a_{k-1},a_{k}$ are the two terms "in the middle" of the seqeunce. Group 
$a_1$ with $a_n$, group $a_2$ with $a_{n-1}$, group $a_3$ with $a_{n-2}$,... group $a_{k-2}$ with $a_{k+1}$, and group $a_{k-1}$ with $a_k$. We have that 
$a_2a_{n-1}\ge a_1a_n$, $a_3a_{n-2}\ge a_1a_n$, ..., $a_{k-2}a_{k+1}\ge a_1a_n$, $a_{k-1}a_k\ge a_1a_n$, and trivially $a_1a_n\ge a_1a_n$. Since there are exactly $n/2$ such groups,we obtain $a_1a_2a_3...a_{k-2}a_{k-1}a_{k}a_{k+1}...a_{n-2}a_{n-1}a_n \ge (a_1a_n)^{n/2}$. The case when $n$ is odd is similar, left to you as an exercise. 
