Inequality between two functions I have a two functions defined for $x > 1$, and $c \in (0,1)$:
$$
f(x) = 1-\exp\left(-\frac{c}{x^2} \right),
$$
and
$$
g(x) = \exp\left(-\frac{x}{c} \right).
$$
From graphical tool (
https://www.desmos.com/calculator/hr8n8kkpym ), I know $f(x) > g(x)$. How can I prove this inequality analytically? 
 A: Calling
$$
\left\{ \begin{array}{rcl}
u & = & 1-e^{-\frac{c}{x^{2}}}\\
v & = & e^{-\frac{x}{c}}
\end{array}\right. (1)
$$
we have
$$
\left\{ \begin{array}{rcl}
\log(u) & = & \log(1-e^{-\frac{c}{x^{2}}})\\
\log(v) & = & -\frac{x}{c}
\end{array}\right. (2)
$$
and also
$$
\left\{ \begin{array}{rcl}
\frac{d}{dx}\log(u) & = & -\frac{1}{(e^{\frac{c}{x^{2}}}-1)x^{3}}\\
\frac{d}{dx}\log(u) & = & -\frac{1}{c}
\end{array}\right. (3)
$$
and for $x=1$ we have
$$
-\frac{1}{e^{c}-1}>-\frac{1}{c}
$$
and for $x>1$ and $c\in(0,1)$ we have
$$
-\frac{1}{(e^{\frac{c}{x^{2}}}-1)x^{3}}>-\frac{1}{c}
$$
Now ressuming, by (3) if $\log(1-e^{-c})>-\frac{1}{c}$ then $\log(u)$
and $\log(u)$ does not intersect and will remain $\log(u)>\log(v)$
all along $x\ge1$
Concluding, $\log$ is a strict monotonic increasing function for
$x>1$ hence $u > v$
A: As x tends to infinity f(x) tends to 0 in a slower rate which can be seen by differentiation compared to g(x) 
You can prove it by finding inverse function of f and g which are simpler to prove than this
