Is this proof valid to prove that $F(x)-G(x)$ is always non zero? I have two functions:
$F(x)=16 \left(4 \pi  x^2+\pi \right)^2 x^4+2$
$G(x)=4 x \left(x \left(96 x^4+104 x^2+\left(4 \pi  x^2+\pi \right)^2+2\right)-2 \pi  \left(4 x^2+1\right)^3 \coth (\pi  x)\right)+\left(384 x^6-96 x^4+8 x^2-2\right) \cosh (2 \pi  x)$
I want to show that $F(x)-G(x)$ can never be zero for any value of  $0<x<0.5$. I assume that it can be zero, i.e. $F(x)=G(x)$ for all values of $x$, and try to find a contradiction. Then, we have:
$F(0)=2\quad$  and,  $\quad\lim_{x\to 0} \, G(x)=-10$
Therefore, as I had supposed that $F(x)=G(x)$ should be held for all values of $x$ and now I have found a counterexample, so, the assumption is false and $F(x)-G(x)$ can never be zero for any value of $x$. Is this claim true? If not, does anyone have an idea or hint?
Actually $F$ is always positive, and $G$ is always negative, but it is not easy to prove the latter one.
I thank anyone in advance for their help.
 A: If you have indeed shown that $F(0) - G(0) \neq 0$, this still does not rule out the possibility that $F\left(\frac14\right) - G\left(\frac14\right) = 0$
or $F(0.15) - G(1.5) = 0$ or even
$F\left(\frac1\pi\right) - G\left(\frac1\pi\right) = 0,$
any one of which would be a counterexample to the claim that
"$F(x) - G(x)$ can never be zero for any value of  $0<x<0.5$."
The problem with your "proof" is that you're acting as if "always zero" is the opposite of "never zero." The opposite of "never zero" is "sometimes zero".
More generally, a claim like "$F(x) - G(x)$ can never be zero"
is disproved if you find even one counterexample, that is, for a disproof all you need is one value of $x$ where $F(x) - G(x) = 0.$
If something can be disproved by a single counterexample,
you're usually not going to be able to prove it by a single counterexample.
But if you know that $F(x) > 0$ for every $0 < x < 0.5$, and you also know that
$G(x) < 0$ for every $0 < x < 0.5$,
then you know  $-G(x) > 0$ for every $0 < x < 0.5$
(because the negation of a negative number is a positive number)
and therefore $F(x) - G(x) > 0$ for every $0 < x < 0.5$
(because $F(x) - G(x) = F(x) + (-G(x))$ and the sum of two positive numbers is positive).
