I would like to show that $$(m+1)\,c_m- \sum_{n=1}^{m-1} c_n \,\sigma_{m-n} > 0$$ for all $m$, where $c_i$ is the $i$th coefficient of Klein's $J$-function $$J(q)= \frac{1728 \;E_4^3(q) }{E_4^3(q)-E_6^2(q)}-744$$ (in terms of Eisenstein series $E_4$ and $E_6$) and $\sigma_i$ is minus the $i$th coefficient of the second order Eisenstein series $E_2$. Both are positive numbers, bounded respectively by $$c_n \leqslant \frac{e^{4\pi\sqrt{n}}}{\sqrt{2}\;n^{3/4}}$$ $$c_n \geqslant \frac{e^{4\pi\sqrt{n}}}{\sqrt{2}\;n^{3/4}}\, K$$ $$\sigma_n < 24 \left[e^{\gamma}\,n\,\text{ln}(\text{ln}(n))+\frac{0.6483\, n}{\text{ln}(\text{ln}(n))}\right]\,\qquad (n\geqslant 3)$$ where $\gamma$ is Euler's constant ($\sim 0.6$) and $K\sim 0.98$. I could also use the cruder bound $$c_n \leqslant e^{4\pi \sqrt{n+1}}\,.$$ I've tried bounding the quotient of the two terms, Taylor-expanding and taking a limit to show that it goes to zero, but that's only valid for small values of $n$ and I'm interested in all kinds of values. Numerical evidence suggests this quotient goes to something like 1, but I don't know how to show that, if I must refrain from Taylor-expanding. (Also, replacing the sum by $(m-1)c_{m-1}\sigma_1$ is much too brutal).

Any ideas are welcome, and if there's any interesting information on the coefficients that I'm not aware of, I would appreciate also! For example, knowledge of the relations between successive coefficients might enable one to make a recursive type of argument, but I haven't found any such thing.

Edit: To be more complete and answer David Loeffler's comment, this expression appears in the modular form of weight 2 $$q\partial_q\,J(q)+E_2\left(J(q)+24\right)=\sum_{m=1}^{\infty}F(m)\,q^m$$ with $F(m)=(m+1)\,c_m- \sum_{n=1}^{m-1} c_n \,\sigma_{m-n}-\sigma_{m+1}-\sigma_1\sigma_m$ (I'm assuming the sigma terms are negligible).