How prove this inequality $(\sum a_{1}^{1.5})^2\ge \sum a_{1}\sum a_{1}a_{2}$ Now my question let $a_{1},a_{2},\cdots,a_{n}$ are positive numbers,and $a_{n+i}=a_{i},i=1,2,\cdots$,show that
$$(\sum a_{1}^{1.5})^2\ge \sum a_{1}\sum a_{1}a_{2}$$
my teacher (tian275461) have prove this
$$(a^{1.5}+b^{1.5}+c^{1.5})^2\ge (a+b+c)(ab+bc+ac)$$
He methods:let $a\longrightarrow a^2,b\longrightarrow b^2,c\longrightarrow c^2$
then
$$\Longleftrightarrow (a^3+b^3+c^3)^2\ge (a^2+b^2+c^2)(a^2b^2+c^2a^2+b^2c^2)$$
$$\Longleftrightarrow(\sum a^3)^2\ge \sum a^2\sum a^2b^2$$
note
$$(\sum a^3)^2=\sum a^2\sum a^4-\sum b^2c^2(b-c)^2$$
$$\Longleftrightarrow \sum a^2\left(\sum a^4-\sum a^2b^2\right)-\sum b^2c^2(b-c)^2\ge 0 $$
$$\Longleftrightarrow \dfrac{1}{2}\sum a^2\sum(b^2-c^2)^2-\sum b^2c^2(b-c)^2\ge 0$$
$$\Longleftrightarrow \dfrac{1}{2}(b-c)^2 \left(\sum a^2\sum (b+c)^2-2\sum b^2c^2\right) \ge 0$$
it suffices to show that
$$\sum b^2\sum (b+c)^2-2\sum b^2c^2\ge 0$$
and note that
$$\sum b^2\sum (b+c)^2-2\sum b^2c^2=2\sum a^4+2\sum a^3b+2\sum a^3c+2\sum a^2b^2+2\sum a^2bc\ge 0$$
for n=4,it only show that
$$(a^3+b^3+c^3+d^3)^2\ge (a^2+b^2+c^2+d^2)(a^2b^2+b^2c^2+c^2d^2+d^2a^2)$$
 A: The conjecture is false.
(See my previous post for sufficient extra conditions such that the conjecture holds.)
Here is a counterexample. Let n even (n odd works as well).
Let $a_1=a_2 = \dots = a_{n/2} = A$ and $a_{n/2 +1 } = a_{n/2 +2 } = \dots = a_{n} = B$. Let, without loss of generality, $B<A$. 
Then the conjecture is
$$
(\sum a_{1}^{1.5})^2 -  \sum a_{1}\sum a_{1}a_{2} = \\
\frac{n^2}{4}(A^{1.5} +B^{1.5})^2 - \frac{n}2 (A+B) ((\frac{n}2-1)(A^2+B^2) + 2 AB) \ge 0
$$
Let $B = x^2\cdot A$, (where $0<x<1$), then this transforms into
$$
n(1 +x^{3})^2 - (1+x^2) ((n-2)(1+x^4) + 4 x^2) \ge 0
$$
Expanding this gives
$$
n(2x^3 - x^2 - x^4) - 2 x^2 - 2 x^4 + 2 x^6 + 2 \ge 0 
$$
Now the factor of the leading $n$ is always negative: $2x^3 - x^2 - x^4= -( x - x^2)^2 < 0 $ for $0<x<1$. For the remaining sum terms, $2 > - 2 x^2 - 2 x^4 + 2 x^6 + 2 >0$ for $0<x<1$. 
Hence the conjecture fails for large enough $n$, namely 
$$
n  > n_0 = \frac{- 2 x^2 - 2 x^4 + 2 x^6 + 2 }{- (2x^3 - x^2 - x^4) }
$$
$
\Box
$
The following plots illustrate the magnitude of $n_0$ for two ranges of $x$. From those plots, one could conjecture, if my example is indeed the worst case,  that the proposed inequality  holds for all $n<16$. ($n=3$ and $n=4$ are proven.)


A: Here is a proof in the case where the $\{a_i\}$ obey a condition for the variance about their mean.
More precisely, the formulation for the above condition is the following.  Let $\bf a$  be the vector composed from the $\{a_i\}$, i.e. ${\bf a} = (a_1,a_2, \dots, a_{n-1}, a_n)$. Define the mean ${m}  = \frac1n \sum a_i$. Define a vector ${\bf m} = (m,m, \dots, m)$. Let $\bf \bar a$  be the permutation of $\bf a$ which is constructed by shifting all components cyclically by one position, i.e.  ${\bf \bar a} = (a_2,a_3, \dots, a_n, a_1)$. Then we require 
$$
\cos^2({\bf a}, {\bf m})\geq \cos({\bf a}, {\bf \bar a}) 
$$
Writing this in components gives the equivalent
$$
\frac{m^2 (\sum_j a_j)^2}{n m^2 \sum_j a_j^2 } \geq \frac{\sum_j a_j a_{j+1}}{\sum_j a_j^2 }
$$
or $(\sum_j a_j)^2 \ge n  \sum_j a_j a_{j+1}$. Alternatively, one can define the difference $d_j = a_j - m$. Then we have the condition $\sum_j d_j d_{j+1} < 0$.
As an example, consider ${\bf a} = (10,9,9,1)$. Then we have 
 $(\sum_j a_j)^2 = (29)^2 = 841 \ge n  \sum_j a_j a_{j+1} = 4 \cdot(90 + 81 + 9 + 10) = 760$. 
As a counterexample, consider ${\bf a} = (4,5,4,2,1,2)$. Then we have 
 $(\sum_j a_j)^2 = (18)^2 = 324 < n  \sum_j a_j a_{j+1} = 6 \cdot(20 + 20 + 8 + 2 + 2 +8) = 360$.
(Note that, nevertheless, the original inequality holds in this case.)
Now it suffices to prove, instead of the original inequality, the tighter one:
$$
n (\sum_j a_j^{1.5})^2 \ge (\sum_j a_j)^3
$$
Setting $a_j = b_j^2$ gives
$$
 (\frac1n \sum_j b_j^{3})^2 \ge (\frac1n \sum_j b_j^2)^3
$$
or 
$$
 \sqrt[3]{\frac1n \sum_j b_j^{3}} \ge \sqrt[2]{\frac1n \sum_j b_j^{2}} 
$$
but this is just the power mean inequality.$\qquad \Box$ 
