I had a similar problem posted here, and after experimentation with WA I noticed that the function $1-\left(\frac{\sin x}{x}\right) ^{\frac{2}{5}}$ has a Maclaurin expansion with all coefficients positive.
It seems to hold. A possible approach might using the product expansion for the function $\sin x$. Or maybe a differential equation.
Thank you for your interest!
$\bf{Added:}$ Denote the expression by $y(x)$. It's enough to show that $z\colon=y'$ has a positive expansion. We have
$$x\cdot z' = \left(\frac{x^2}{1 – x \cot x} + \frac{3}{5}( 1- x \cot x) – 2 \right)\cdot z$$
If the expression in the brackets has a positive expansion (it seems so) then we can show that $z$ has a positive expasion.
Best Answer
Not an answer, but here's my attempt: the function $f(x) = 1 - \left(\frac{\sin x}{x}\right)^{2/5}$ satisfies a differential equation $$ \frac{y'}{1-y} = -\frac{2}{5x}(x\cot x - 1) $$ so the coefficients of the series $f(x) = 1 + \sum_{k\geq 1} a_{2k}x^{2k}$ ($f$ is an even function) satisfy the recurrence relation $$ 2na_{2n} = \frac{2}{5}\left(b_{2n} - a_{2}b_{2n-2} - a_{4}b_{2n-4} - \cdots - a_{2n-2}b_{2} \right) $$ where $$ b_{2k} = \frac{(-1)^{k+1}B_{2k}2^{2k}}{(2k)!} $$ and $B_{2k}$ is Bernoulli number. Note that $b_{2k} > 0$ for all $k\geq 1$. Assume that the following inequality holds for all $n\geq 2$: $$ \sum_{1\leq k \leq n-1} \frac{1}{5k}b_{k}b_{n-k} \leq b_{n} $$ then one can prove $0\leq a_{n} \leq \frac{b_{2n}}{5n}$ for all $n$, by induction on $n$. However, I failed to prove the above inequaltiy, and I don't even know whether the inequality is true or not. I tried to use an identity $$ \sum_{1\leq k \leq n} b_{k}b_{n-k} = (2n+1)b_{n} $$ (for $n\geq 2$) which comes from $$ \coth x = \sum_{n \geq 0} \frac{2^{2n}B_{2n}x^{2n-1}}{(2n)!} $$ and $\frac{d}{dx}\coth x = 1 - \coth^{2}x$, but failed.