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 $$(a^3+b^3+c^3+d^3)^2\ge (a^2+b^2+c^2+d^2)(a^2b^2+b^2c^2+c^2d^2+d^2a^2)$$

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.)

I've checked the OP's equivalence relations and there is a problem due to sloppy sum notation. Here I added what is precisely summed up. Summing over $(b,c)$ stands for the $N(N-1)/2$ many pairs $(b,c)$ where $b \neq c$, when the summand is symmetric under exchange of $b,c$. For additional clarity, I added how many terms each sum encompasses, in [brackets].

This gives a proof for $N=3$, and it also shows why the OP's approach will not directly give results for $N>3$.

The transformation steps are:

$$(\sum_{a \; [N]} a^3)^2=\sum_{a \; [N]} a^6 + 2 \sum_{(b,c) \; [N(N-1)/2]} b^3c^3= \sum_{a \; [N]} a^2\sum_{a \; [N]} a^4-\sum_{(b,c) \; [N(N-1)/2]} b^2c^2(b-c)^2 $$

Using this, the original question transforms into the equivalent $$ \sum_{a \; [N]} a^2\left(\sum_{a \; [N]} a^4-\sum_{a_1 \; [N]} a_1^2 a_2^2\right)-\sum_{(b,c) \; [N(N-1)/2]} b^2c^2(b-c)^2\ge 0 $$ Note in this term, I have carefully used the notation $a_1 a_2$ which generates only $N$ many terms (as required in the original task), as compared to $(a,b)$ which would generate $N(N-1)/2$ many terms. $$ \Longleftrightarrow \dfrac{1}{2}\sum_{a \; [N]} a^2\sum_{a_1 \; [N]}(a_1^2-a_2^2)^2-\sum_{(b,c) \; [N(N-1)/2]} b^2c^2(b-c)^2\ge 0 $$ A slight change leads to $$ \Longleftrightarrow \dfrac{1}{2}\sum_{a_1 \; [N]}(a_1-a_2)^2(a_1+a_2)^2 \sum_{a \; [N]} a^2-\frac12 \sum_{a_1 \; [N]} \sum_{c\neq a_1 \; [N-1]} a_1^2c^2(a_1-c)^2\ge 0 $$ and splitting off $a_2$: $$ \Longleftrightarrow \sum_{a_1 \; [N]}(a_1-a_2)^2(a_1+a_2)^2 \sum_{a \; [N]} a^2-\sum_{a_1 \; [N]} a_1^2a_2^2(a_1-a_2)^2\\ -\sum_{a_1 \; [N]} \sum_{c\neq a_1, a_2 \; [N-2]} a_1^2c^2(a_1-c)^2\ge 0 $$ which simplifies to $$ \Longleftrightarrow \sum_{a_1 \; [N]}(a_1-a_2)^2 {\Large [} (a_1+a_2)^2 \sum_{a \; [N]} a^2- a_1^2a_2^2 {\Large ]} \\ -\sum_{a_1 \; [N]} \sum_{c\neq a_1, a_2 \; [N-2]} a_1^2c^2(a_1-c)^2\ge 0 $$

At this point it becomes clear that this result will be difficult to use for general $N$.

For $N=3$, the last term is easily modified:

$$ \sum_{a_1 \; [N]} \sum_{c\neq a_1, a_2 \; [N-2]} a_1^2c^2(a_1-c)^2 = \sum_{a_1 } a_1^2 a_3^2(a_1-a_3)^2 = \sum_{a_1 } a_1^2 a_2^2(a_1-a_2)^2 $$ and we have to show $$ \sum_{a_1 }(a_1-a_2)^2 {\Large [} (a_1+a_2)^2 \sum_{a } a^2- 2 a_1^2a_2^2 {\Large ]} \ge 0 $$ which is evident since all factors $$ (a_1+a_2)^2 \sum_{a } a^2- 2 a_1^2a_2^2 \ge (a_1^2+a_2^2)(a_1^2+a_2^2)- 2 a_1^2a_2^2 = a_1^4+a_2^4\ge 0 $$

For $N=4$, the last term gives $$ \sum_{a_1 \; [N]} \sum_{c\neq a_1, a_2 \; [N-2]} a_1^2c^2(a_1-c)^2 = \sum_{a_1 } {\Large [} a_1^2 a_2^2(a_1-a_2)^2 + a_1^2 a_3^2(a_1-a_3)^2 {\Large ]} $$ so including this term into the previous summations will not be directly possible.