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.

While this is just a partial answer, I hope this serves at least as a step in the right direction for proving what you need to.  
First, to work with something more concrete, I substituted the expressions for $f_n(x)$ into $F_m(x)$ and then that into $G_m(x)$ in order to get an explicit set of functions:
$$F_m(x) = \sum\limits_{n=1}^mn^2x(2\pi n^2x - 3)\exp(-n^2\pi x)$$
$$G_m(x) = \sum\limits_{n=1}^mn^2x(2\pi n^2x - 3)\exp(-n^2\pi x) + \sum\limits_{n=1}^m\frac{n^2}{x}\left(\frac{2\pi n^2}{x} - 3\right)\exp\left(-\frac{n^2\pi}{x}\right)$$
While this class of functions is particularly nasty, we can make some ways with proving the proposed bounds $m + 1 > x_0(m) > m + 2$ hold. (By the way, the inequalities are reversed here than in your question, since the function $G_m(x)$ is positive and decreases in $x_0(m)$'s neighborhood, so $G_m(m + 1) > G_m(m + 2)$, thus making the former the upper bound and the latter the lower bound, not vice versa.)  
So basically what we need to show is that $G_m(m + 1)$ is always positive for all $m$, and $G_m(m + 2)$ is always negative, thus by the intermediate value theorem (since the function in question is continuous), $G_m(x)$ has a root in between $m + 1$ and $m + 2$.  
Substituting in $x = m + 1$ for the function, we get:
$$G_m(m + 1) = \sum\limits_{n=1}^mn^2(m + 1)(2\pi n^2m + 2\pi n^2 - 3)\exp(-n^2\pi (m + 1)) + \sum\limits_{n=1}^m\frac{n^2}{m + 1}\left(\frac{2\pi n^2}{m + 1} - 3\right)\exp\left(-\frac{n^2\pi}{m + 1}\right)$$
If we can show that each factor in each summation is always positive, then the whole summation is positive and so the functions are always positive at that point. (Actually, that would be the best case scenario of it satisfying a sufficient but not necessary condition; it can have some negative sums as long as the total value of the positive terms is larger than the negative ones.)  
From the first summation, factor by factor:  
1st summation, 1st factor: $n^2$
Since the square of any number is always positive, $n^2$ is positive.  
1st summation, 2nd factor: $m + 1$
Obviously $m$ is a positive number by definition, and so $m + 1$ is also always positive.  
1st summation, 3rd product: $(2\pi n^2m + 2\pi n^2 - 3)$
$2\pi n^2m + 2\pi n^2$ must be greater than 3 for this factor to be positive. Taking the 'lowest' case of $n = m = 1$, we get $2\pi + 2\pi = 4\pi,$ and $4\pi > 3$, so this factor will always be positive.  
1st summation, 4th product: $\exp(-n^2\pi (m + 1))$
An exponential term is never negative or zero.  
Now, we go on to the second summation:
2nd summation, 1st product: $\frac{n^2}{m + 1}$
$n^2$ and $m + 1$ are always positive, so their quotient is too.  
2nd summation, 2nd product: $\frac{2\pi n^2}{m + 1} - 3$
Alas, here we run into trouble. Taking the case $n = 1, m = 2$, we see that the resulting term is negative. When will it be negative? Like the corresponding factor in the first summation, the fraction must be greater than $3$:  
$\frac{2\pi n^2}{m + 1} > 3 \implies 2\pi n^2 > 3(m + 1) \implies n > \sqrt{\frac{3}{2\pi}}\sqrt{m + 1}$
So, for the first few terms in the sum, the product will be negative, but once n is sufficiently large to satisfy the inequality, the product will become positive.
2nd summation, 3rd product: $\exp\left(-\frac{n^2\pi}{m + 1}\right)$
Again, an exponential term is never negative.  
So, what does this all mean since not all terms are positive? All it means is that we need to prove that the first summation is larger than the second one (which I couldn't do), so that their difference is still positive:
$$\sum\limits_{n=1}^mn^2(m + 1)(2\pi n^2m + 2\pi n^2 - 3)\exp(-n^2\pi (m + 1)) > \sum\limits_{n=1}^m\frac{n^2}{m + 1}\left(\frac{2\pi n^2}{m + 1} - 3\right)\exp\left(-\frac{n^2\pi}{m + 1}\right)$$
Or, if we strip out the positive terms from the second summation, (I'm calling the terms $a_n$ and $b_n$ for the 1st and 2nd summations respectively)
$$\left[\sum\limits_{n=1}^ma_n + \sum\limits_{n > \sqrt{\frac{3}{2\pi}}\sqrt{m + 1}}^mb_n\right] > \sum\limits_{n < \sqrt{\frac{3}{2\pi}}\sqrt{m + 1}}^mb_n$$
If we prove either inequality (the first one is stronger than the second), we deduce that the function is always positive at the point $m + 1$. We can use an extremely similar argument for the point $m + 2$ to prove it is negative, and thus we will have proved the bounds. About the first question (whether $x_0(m)$ is the smallest zero of $G_m(x)$), if we take that the function is decreasing from $G_m(1)$ to $x_0(m)$, we can prove it to be the smallest zero by contradiction (and if we also accept that $\lim\limits_{x \to \infty} G_m(x) = 0$, then we can prove that it is the only zero.) For suppose that there exists other zeros smaller than $x_0(m)$; that is, in the interval $[1, x_0(m)]$. Since the function is continuous, the only way for it to have a smaller zero is if the function dips below zero and back up again (since it needs to pass through zero at $x_0(m)$). But since the function is always decreasing on that interval, then we reach a contradiction, since after the first 'smaller' zero the function would be negative and would need to be increasing to cross the x-axis again. I know this is far from rigorous, but either way proving the bounds would also prove this.  
I apologize for the (extremely!) long answer, but I found out a lot about this function and didn't want anything to go to waste. Cheers!
