Is $\sum_i c_i (x^{n_i} + x^{-n_i} ) \ge 0$ for $\sum_i c_i =0$ and $\sum_i c_i n_i^2 > 0$? Let $ f(x) = \sum_i c_i (x^{n_i} + x^{-n_i} )$ for an index set $\{i\}$ with $n_i \in \cal{{N}}_0$; $\sum_i c_i = 0$, and $c_i >0 $ for the maximum $n_i$.
Further, it is required that   $\sum_i c_i n_i^2 >  0$, which ensures that a local minimum is observed at  $x=1$.
Then prove or disprove that $f(x) \ge 0$ for $x > 0$, where equality occurs only  for $x=1$.
Some examples for explanation:
$f_1(x) = x^2 + x^{-2} -2$ satisfies the conditions with $c_1 = 1, n_1 = 2$ and $c_2 =-1, n_2 = 0$. Indeed, $f_1(x) = (x^1 - x^{-1})^2 \ge 0$ with equality only for $x = 1$.
$f_2(x) = x^4 + x^{-4} -3  x^2 -3 x^{-2} + 2 x + 2 x^{-1}$ satisfies the conditions with $c_1 = 1, n_1 = 4$ ; $c_2 = -3, n_2 = 2$ and $c_3 =2, n_3 = 1$. Indeed, $f_2(x) =(x - 1)^2 (x^6 + 2 x^5 + 2 x + 1)/x^4 \ge 0$ with equality only for $x = 1$.
A number of other trials showed no counterexamples.
Descartes rule of signs gives only a partial result since there can be numerous changes in the signs of the $\{c_i\}$.

A counterexample:
$$
\begin{align}
 f(x) &= (x^3+x^{-3}) - 50 \cdot(x^2+x^{-2}) + 200 \cdot(x + x^{-1}) - 151\cdot 2 \\
 &= \frac{(x-1)^2(x^4-48 x^3+103 x^2-48 x+1)}{x^3}
\end{align}
$$
satisfies the given conditions, but $f(2) = -51/8 < 0$.
Remark: This was inspired by the representation
$$\sum_ic_i\cosh tn_i=\frac{t^2}2\sum_ic_in_i^2+\frac{t^4}{24}\sum_ic_in_i^4+\cdots
$$
suggested by @TheSimpliFire in a comment to the question. While trying to prove or disprove that
$$
 c_N > 0, \sum_{n=1}^N c_n = 0, \sum_{n=1}^N n^2 c_n > 0 \implies \sum_{n=1}^N n^4 c_n > 0
$$
I found the counterexample $(c_0, c_1, c_2, c_3) = (-151, 200, -50, 1)$ which satisfies
$$
 c_0 + c_1 + c_2 + c_3 = 0 \, ,\\
 c_1 + 4 c_2 + 9 c_3 = 9 > 0 \, , \\
 c_1 + 16 c_2 + 81 c_3 = -519 < 0 \, . \\
$$
Luckily, that lead also to a counterexample for the original problem. More counterexamples can found in a similar way.
