A particular case of the Jacobi's triple product with an inequality and hyperbolic/trigonometric function

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Problem/conjecture :

Show for $x\in[0,1]$ that :

$$\frac{x}{\sin(x)}\geq \cosh\left(\frac{1}{\sqrt{3}}x\right)+\cosh\left(\frac{1}{6}x^2\right)-1+\cosh\left(\frac{1}{27}x^4\right)-1+\sum_{n=3}^{\infty}\left(\cosh\left(\frac{2^n}{5^n}x^{n^2}\right)-1\right)\tag{I}$$

I can show the weaker inequality $x\in[0,1]$ then:

$$\frac{x}{\sin(x)}\geq \cosh\left(\frac{1}{\sqrt{3}}x\right)$$

Using $x\in[0,1]$:

$$\frac{x}{\sin(x)}\geq \frac{\sinh(x)}{x}\geq \cosh\left(\frac{1}{\sqrt{3}}x\right)$$

Question :

Have you a sketch to show or a counter-example to (dis)prove $(I)$ ?

My motivation comes from https://en.wikipedia.org/wiki/Jacobi_triple_product#:~:text=In%20mathematics%2C%20the%20Jacobi%20triple%20productis%20the%20mathematical,y%2C%20with%20%7Cx%7C%20%3C%201%20and%20y%E2%89%A0%200.

Any helps is greatly appreciated

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Using series expansion, it is not very difficult to show that $$\text{lhs}~-~\text{rhs}=\frac{x^4}{1080}\,\left( 1+\frac{136 }{63}x^2+O\left(x^4\right)\right)$$ which is already obtained ignoring the summation which does not contribute too much since $$\sum_{n=3}^{\infty}\left(\cosh\left(\frac{2^n}{5^n}x^{n^2}\right)-1\right)=\frac{32 \,x^{18}}{15625}\left(1+\frac 4{25}x^{14}+O\left(x^{18}\right) \right)$$

Numerically, the maximum of the difference is about $ 1.14873\times 10^{-3}$ which takes place around $x=0.912083$.