When does it stop being true?

For $A(a_{1},a_2,\cdots,a_{n})=a_{1}^{2}-\frac{1}{a_{1}^{2}}+a_{2}^{2}-\frac{1}{a_{2}^{2}}\cdots+a_{n}^2-\frac{1}{a_{n}^{2}}$ to have a positive value, the maximum has to be positive.

This is a constrained optimization problem, with a Lagrangian of $\large \mathcal{L}(a_{1},a_2,\cdots,a_{n})=a_{1}^{2}-\frac{1}{a_{1}^{2}}+a_{2}^{2}-\frac{1}{a_{2}^{2}}\cdots+a_{n}^2-\frac{1}{a_{n}^{2}}-\lambda\cdot(a_{1}+a_{2}+\cdots+a_{n}-n)$ . Setting $\large \nabla\mathcal{L}=0$ :

$\large a_{1}+a_{2}+\cdots+a_{n}=n$

$\large a_1+\frac{1}{a_1^3}=a_2+\frac{1}{a_2^3}=\cdots= a_n+\frac{1}{a_n^3}=\lambda$

Since the function $f(x)=x+\frac{1}{x^3}$ has 2 $x$ 's, one on either side of $1$ , for each value of $f>\frac{4}{3^{\frac{3}{4}}}$ and 0 for $f<\frac{4}{3^{\frac{3}{4}}}$ , we can set $a_1=a>1$ and $a_2=a_3=\cdots=a_n=\frac{n-a}{n-1}$ . (The alternative is $a_1=a_2=\cdots=a_n=1$ , which is clearly not a solution.)

We can then find the $a$ 's which optimize $A$ 's value for each $n$ , since $\large a+\frac{1}{a^3}=\frac{n-a}{n-1}+\frac{(n-1)^3}{(n-a)^3}$ has to be true. Numerically solving for each $n$ and evaluating $A(a_{1},a_2,\cdots,a_{n})$ , we have $A$ 's maximum reach a positive value for the first time for $n=11, a_1\approx 4.79264,a_2=a_3=\cdots=a_n\approx 0.620736$ , with $A\approx 0.82608$ , making the answer $\textbf{n=11}$ .