Is it true that $a_1^2a_2^2+a_2^2a_3^2+\cdots+a_8^2a_1^2 \leq 8$ if $a_1^3+a_2^3+\cdots+a_8^3=8 $ for positive numbers? As in the title

Is it true that $a_1^2a_2^2+a_2^2a_3^2+\cdots+a_8^2a_1^2 \leq 8$ if $a_1^3+a_2^3+\cdots+a_8^3=8    $  for positive numbers?

I'm not sure how to answer it. I've got some conclusion but those do not help, so any help would be appreciated.
 A: The inequality does not hold. Try
$$a = (0.150104,0.560857,1.0906,1.429,1.37785,0.967096,0.437358,0.0989485).$$
I obtained this vector as one of the candidates to min/max coming from the Lagrange multipliers method.
A: The basic idea is that big jumps are bad - in particular jumping from above 1 to below 1.
The simplest way to achieve this is $(1,x,x,x,1,y,y,y)$ where $x^3+y^3=2$ and $x\neq 1$. We need to maximize $2x^4+2x^2 + 2y^4+2y^2$. By AMGM, when $x\neq 1$, $2x^4+2x^2> 4x^3$, so $2x^4+2x^2 + 2y^4+2y^2>4x^3+4y^3=8$ giving the desired elementary counterexample.
A: Let
$$f(x,y,z)=x^2y^2+y^2z^2+z^2x^2,\tag1$$
$$F(\overrightarrow a) = a_1^2a_2^2+a_2^2a_3^2+a_3^2a_4^2+\dots+a_8^2a_1^2,$$
then
$$F(\overrightarrow a) = f(a_8,a_1,a_2) + f(a_2,a_3,a_4) + f(a_4,a_5,a_6) + f(a_6,a_7,a_8) - (a_2^2+a_6^2)(a_4^2+a_8^2).$$
If $$a_4=a_8\to 0,\quad a_2=a_6 = p,\quad a_1=a_3=a_5=a_7=q,\tag2$$
then
$$F(\overrightarrow a) = 4p^2q^2\bigg|_{p^3+2q^3=4}=4\sqrt[3]4\,\left(q\sqrt[3]{2-q^3}\right)^2 = g(q),$$
where
$$g' = 8\sqrt[3]4q\left(\sqrt[3]{2-q^3}\right)^2\left(1-\dfrac{q^3}{2-q^3}\right) , $$
$$\max F(\overrightarrow a) = 8\sqrt[3]2\quad\text{at}\quad q=1,\quad p=\sqrt[\large3]2.\tag3$$
Therefore, the OP bound should be increased.
A: A possible approach:
First, note that
$$\left(\frac{\sum a_i^2 a_{i+1}^2} { 8}\right) ^{1/2} \le \left(\frac{\sum a_i^3 a_{i+1}^3} { 8}\right) ^{1/3}$$
Now, it is enough to check that
$$\sum_{i=1}^8 a_i^3 a_{i+1}^3 \le 8$$
if $a_i$ are positive with  $\sum a_i^3 =1$, or, equivalently
$$\sum_{i=1}^8 b_i b_{i+1} \le 8$$
if $b_i$ are positive with sum $8$.  Now, it is enough to check that
$$ \frac{\sum_{i=1}^ 8  x_i x_{i+1} }{ 8 } \le \left ( \frac{\sum x_i}{8} \right )^2 $$
for $x_i$ positive ( and say $\le 1$) .
$Added:$  It turns out that the last inequality is not correct, as @Martin R: pointed out.  It may work for some smaller values of $n$, but not $n=8$.
Now it seems that the original inequality is also not correct.
