Upper and lower bounds for the smallest zero of a function The function $G_m(x)$ is what I encountered during my search for approximates of Riemann $\zeta$ function:
$$f_n(x)=n^2 x\left(2\pi n^2 x-3 \right)\exp\left(-\pi n^2 x\right)\text{,   }x\ge1;n=1,2,3,\cdots,\tag{1}$$
$$F_m(x)=\sum_{n=1}^{m}f_n(x)\text{,  }\tag{2}$$
$$G_m(x)=F_m(x)+F_m(1/x)\text{,  }\tag{3}$$
Numerical results showed that $G_m(x)$ is zero near $m+1.2$ for $m=1,2,...,8$.
Please refer to fig 1 below for the plot of $\log|G_m(x)|$ v.s. $x$ for $m=1,2,...,8$

Let us denote these zeros by $x_0(m)$.
I am interested if it can be proved that
(A) $x_0(m)$ is the smallest zero of $G_m(x)$ for $x\ge1$ 
(B) there exist bounds $\mu(m),\nu(m)$ such that $0<\mu(m)\le x_0(m)\le \nu(m)$;$\mu(m),\nu(m)\to\infty$, when $m\to\infty$.
Here are the things I tried.
Because $G_m(1)>0$ and $G_m(x)\to F_m(1/x)<0$ when $x\to\infty$, so there exist a zero for $G_m(x)$ between $x=1$ and $x=\infty$.
But I was not able to find the bounds for this zero.
It is tempting to speculate that $x_0(m)$ is the only zero for $G_m(x)$ and $m+1<x_0(m)<m+2$.
The values for $x_0(m), (m=1,2,...,10)$ are given by:
$x_0(1)$=2.24203,
$x_0(2)$=3.21971,
$x_0(3)$=4.21913,
$x_0(4)$=5.22283,
$x_0(5)$=6.22764,
$x_0(6)$=7.23268,
$x_0(7)$=8.23764,
$x_0(8)$=9.24241,
$x_0(9)$=10.2469,
$x_0(10)$=11.2512.

@GeorgeDaccache Nice results!
I just need the space and easy of use in this answer area to convey some thoughts that I have.
First let me define two integers $n_0(m)$ and $n_1(m)$ as
$$ n_0(m) =\lfloor{\sqrt{3/(2\pi)}\sqrt{m + 1}}\rfloor$$
$$ n_1(m) =\lceil{\sqrt{3/(2\pi)}\sqrt{m + 1}}\text{ }\rceil$$
For convenience we also define $a_n(m)$ and $b_n(m)$ as:
$$a_n(m)=n^2(m + 1)(2\pi n^2(m + 1) - 3)>0$$
$$b_n(m)=\frac{n^2}{m + 1}\left(\frac{2\pi n^2}{m + 1} - 3\right)$$
So that
$$b_n(m)\gt 0, m\ge n\gt n_1(m)$$
$$b_n(m)\lt 0, 1\le n\lt n_0(m)$$
Then what we want to prove is the following:
$$\sum\limits_{n=1}^m a_n(m)\exp(-n^2\pi (m + 1)) + \sum\limits_{n=n_1(m)}^m b_n(m)\exp\left(-\frac{n^2\pi}{m + 1}\right) >  \sum\limits_{n=1}^{n_0(m)} (-b_n(m))\exp\left(-\frac{n^2\pi}{m + 1}\right)\tag{1}$$
Since for $A>0$,
$$\exp(-A n^2)>\exp(-A m^2)$$
$$\exp(-A n^2)<\exp(-A * 1^2)=\exp(-A)$$
We can replace the exponential terms by their corresponding limit terms, factor them out of the summation and thus only need to prove the following:
$$\exp(-(n_0(m))^2\pi (m + 1))\sum\limits_{n=1}^{n_0(m)} a_n(m)+\exp(-m^2\pi (m + 1))\sum\limits_{n=n_1(m)}^m a_n(m) + \exp\left(-\frac{m^2\pi}{m + 1}\right)\sum\limits_{n=n_1(m)}^m b_n(m) >  \exp\left(-\frac{1^2\pi}{m + 1}\right)\sum\limits_{n=1}^{n_0(m)} (-b_n(m))\tag{2}$$
The nice thing about (2) is that we can now complete the summation of $n$ in (2).
@GeorgeDaccache, can you continue to work along this direction? Since you already spent so much effort on this problem. I feel that you might just be one step away from providing a complete answer.
Best regards-
mike
EDIT:
Numerical results for $m=10$ showed that the terms associated with $a_n$ are quite tiny, so we can now focus on proving that:
$$\sum\limits_{n=n_1(m)}^m b_n(m)\exp\left(-\frac{n^2\pi}{m + 1}\right) >  \sum\limits_{n=1}^{n_0(m)} (-b_n(m))\exp\left(-\frac{n^2\pi}{m + 1}\right)\tag{3}$$
