Bound on $\exp[-1/x]$

Question

How can I show that $e^{-\frac{1}{x}} \leq \frac{x}{e}$ on the interval $x\in[0,1]$ ?

Any help is much appreciated

Answer 1

The interval should be $(0,1]$. If you set $t=1/x$, the inequality becomes $$ e^{-t}\le \frac{1}{et},\qquad \text{for $t\ge1$} $$ or, as well, $$ et\le e^t,\qquad \text{for $t\ge1$} $$ Now consider $f(t)=e^t-et$ on $[1,\infty)$. Since $f(1)=0$ and $$ f'(t)=e^t-e $$ what can you conclude?

You can also do it by taking logarithms: your inequality is equivalent to $$ -\frac{1}{x}\le \log x-1,\qquad\text{for $0<x\le1$} $$ The function $$ g(x)=\log x+\frac{1}{x}-1 $$ has $g(1)=0$ and its derivative is $$ g'(x)=\frac{1}{x}-\frac{1}{x^2}=\frac{x-1}{x^2} $$