Inequality between two functions I have a two functions defined for $x &gt; 1$, 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 $f(x) &gt; g(x)$. How can I prove this inequality analytically?
Calling

$$
\left\{ \begin{array}{rcl}
u &amp; = &amp; 1-e^{-\frac{c}{x^{2}}}\\
v &amp; = &amp; e^{-\frac{x}{c}}
\end{array}\right. (1)
$$

we have

$$
\left\{ \begin{array}{rcl}
\log(u) &amp; = &amp; \log(1-e^{-\frac{c}{x^{2}}})\\
\log(v) &amp; = &amp; -\frac{x}{c}
\end{array}\right. (2)
$$

and also

$$
\left\{ \begin{array}{rcl}
\frac{d}{dx}\log(u) &amp; = &amp; -\frac{1}{(e^{\frac{c}{x^{2}}}-1)x^{3}}\\
\frac{d}{dx}\log(u) &amp; = &amp; -\frac{1}{c}
\end{array}\right. (3)
$$

and for $x=1$ we have

$$
-\frac{1}{e^{c}-1}&gt;-\frac{1}{c}
$$

and for $x&gt;1$ and $c\in(0,1)$ we have

$$
-\frac{1}{(e^{\frac{c}{x^{2}}}-1)x^{3}}&gt;-\frac{1}{c}
$$

Now ressuming, by (3) if $\log(1-e^{-c})&gt;-\frac{1}{c}$ then $\log(u)$
and $\log(u)$ does not intersect and will remain $\log(u)&gt;\log(v)$
all along $x\ge1$

Concluding, $\log$ is a strict monotonic increasing function for
$x&gt;1$ hence $u &gt; v$
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