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?
On
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 $u > v$
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