The Maclaurin series of $1-(1-\frac{x^2}{2} + \frac{x^4}{24})^{2/3}$ has all coefficients positive It was shown in a previous post that the Maclaurin series of $1 - \cos^{2/3} x$ has positive coefficients. There @Dr. Wolfgang Hintze: has noticed that the truncation $1- \frac{x^2}{2} + \frac{x^4}{24}$ can be substituted for $\cos x$ (  seems to be true for all the truncations). The proof is escaping me.
Thank you for your attention!
$\bf{Added:}$ Thomas Laffey in this paper directs to a proof of the fact that if $a_1$, $\ldots$, $a_n\ge 0$ then $\alpha = \frac{1}{n}$ makes the following series positive:
$$1- (\prod_{i=1}^n (1- a_i x))^{\alpha}$$
Numerical testing suggests that $\alpha = \frac{\sum a_i^2}{(\sum a_i)^2} \ge \frac{1}{n}$ works as well ( see the case $n=2$ tested here). So in our case, instead of $\alpha = \frac{1}{2}$ we can take $\alpha = \frac{2}{3}$. Clearly, this would then be the optimal value. This would be a test case for $n=2$. The result for $\cos x$ used the special properties of the function ( solution of a certain differential equation of second order). Maybe $1- x/2 + x^2/24$ is as general as any quadratic with two positive (distinct) roots.
 A: Hint
Consider
$$f(a)=1-\left(1-\frac{x^2}{2}+\frac{x^4}{24}\right)^a$$ and use the binomial expansion.
It should write
$$f(a)=\frac a 2 \sum_{n=1}^\infty\frac {P_n(a)}{b_n} x^{2n}$$ You should see it very quickly.
Edit
After your edit and remarks, considering
$$f(k)=1-\big[ (1-ax)(1-b x)\big]^k$$  let $b=a c$ to obtain
$$f(k)=k\sum_{n=1}^\infty \frac{a^n}{n!}\, Q_n(c)\, x^n$$ with
$$Q_1(c)=c+1 \qquad\text{and} \qquad Q_2(c)=(c^2+1)-k (c+1)^2$$ If, as you did, we choose to cancel $Q_2(c)$ that is to say to use $k=\frac{c^2+1}{(c+1)^2}$ we have
$$\color{blue}{Q_n(c)=  (n-2)\, 2^{\left\lfloor \frac{n-1}{2}\right\rfloor }\,  \frac{c\,  (c^2-1)^2}{(c+1)^{n}} \,\,R_n(c)}$$ with
$$\left(
\begin{array}{cc}
n & R_n(c) \\
 3 & 1 \\
 4 & c^2+3 c+1 \\
 5 & \left(c^2+c+1\right) \left(c^2+4 c+1\right) \\
 6 & \left(c^2+1\right) \left(c^2+4 c+1\right) \left(3 c^2+8 c+3\right) \\
 7 & \left(c^2+3 c+1\right) \left(c^2+4 c+1\right) \left(6 c^4+7 c^3+6 c^2+7   c+6\right)
\end{array}
\right)$$
A: After testing with WolframAlpha I realized we can prove even more.
First,  it is enough to theck that the derivative of  $f(x) = 1- (1- x/2 + x^2/24)^{2/3}$ has positive coefficients. Even more, it is enough to check that $f'(x) = c \cdot \exp(g(x))$, where $g$ has positive coefficients. But that is equivalent to showing that $\frac{f''(x)}{f'(x)}$ has.  The big advantage here is that this fraction is rational, and is relatively easy to deal with. It is in fact better to consider the general case of
$$f(x) = 1 - ((1-a x)(1- b x))^p$$
where $p$ is an arbitrary exponent. We may also assume $b=1$. We get
$$\frac{f''(x)}{f'(x)} = \frac{1 - p}{1 - x} - \frac{2 a}{1 + a - 2 a x} + \frac{a (1 - p)}{1 - a x}$$
Now for $n\ge 1$ the coefficient of $x^{n-1}$ equals
$$(1-p)(1+a^n) - \frac{(2 a)^n}{(a+1)^n}$$
Now choose $p=\frac{a^2+1}{(a+1)^2}$. We need to show that
$$\frac{2 a}{(1+a)^2}\ge \frac{ (2 a)^n}{(1+a^n)(1+a)^n}$$ for $n\ge 1$ and $a \ge 0$.
Note that for $n=1$ we have equality. (Also note that both sides do not change under $a\mapsto \frac{1}{a}$). To show the inequality for $n\ge 1$ it enough to show that the function
$t \mapsto \frac{ (2 a)^t}{(1+a^t)(1+a)^t}$ is decreasing for $a\ge 1$ on $[1, \infty)$. Its derivative equals
$$\frac{ (\frac {2a}{a+1})^t (\log \frac{2a}{a+1} - a^t \log \frac{a+1}{2})}{(a^t+1)^2}$$
Now observe that $a^t\ge 1$ and $\frac{a+1}{2}\ge \frac{2a}{a+1} \ge 1$, so the top large bracket is negative. We are done.
Note: This method might just work for the general case, but the machine proof would be just for a concrete $n$. In any case, for functions of the form $f(x)= 1- \prod_{i=1}^n(1- a_i x)^{\alpha_i}$ the quotient $\frac{f''(x)}{f'(x)}$ is a rational function.
$\bf{Added:}$ Let's see how the general case might work. Consider a function of the form
$f(x) = 1- g(x)^{\alpha}$. Then we get
$$\frac{f''(x)}{f'(x)} = \frac{(\alpha-1)g'(x)}{g(x)} + \frac{g''(x)}{g'(x)} = \alpha \frac{ g'(x)}{g(x)}+ \frac{g''(x) g(x) - g'(x)^2}{g(x) g'(x)}$$. Note that we have
$$(\frac{g'(x)}{g(x)})'= \frac{ g''(x) g(x) - g'(x)^2}{g(x)^2}$$
In the end we get
$$\frac{f''}{f'} = \alpha \frac{g'}{g} + \frac{(g'/g)'}{g'/g}$$
Now suppose $g(x) = \prod_{i=1}^n (1- a_i x)$. Then $g'/g = - \sum \frac{a_i}{1- a_i x}$ and
$(g'/g)' = -\sum \frac{a_i^2}{(1- a_i x)^2}$ . We get
$$\frac{\sum \frac{a_i^2}{(1-a_i x)^2}}{\sum \frac{a_i}{1-a_i x}}- \alpha \cdot \sum \frac{a_i}{1-a_i x}$$. Each of has coefficients in terms of the Newton sums.
$\bf{Added:}$ In the above expression for $f(x) = 1- g(x)^{\alpha}$ we can iqnore the free term. Therefore, we may consider $g(x)$ any polynomial with real roots and positive free term, say $g(x) = c(r_1-x)(r_2-x) \cdots$. Now $g'(x)/g(x) = - \sum_{k\ge 0} S_{-k-1} x^k$, while $g''(x)/g'(x) = \sum_{k\ge 0} S'_{-k-1} x^k$, where $S_l$, $S'_l$ are the Newton sums for the polynomial $g(x)$, $g'(x)$. Now the optimal $\alpha$ is such that the free term in $(1-\alpha) (-\frac{g'}{g}) + \frac{g''}{g}$ cancels. To show the positivity one needs to show this:
$$\frac{S'_{-k}}{S_{-k}}\le \frac{S'_{-1}}{S_{-1}}$$
for all $k$ natural. With some algebra and the Maclaurin inequalities one can show this for $k=2$ ($n$ arbitrary). Perhaps it is worth thinking about general $k$.
