Inequality $e^{-\tau}\leq e^{-\frac{\tau}{t}}t^{-1}$. Can we prove that
$$e^{-\tau}\leq e^{-\tau/t}t^{-1}$$
for $t \in (0,1)$ and $0<\tau <t$? I have tested this inequality using geogebra, apparently it's true. I have tried to prove that $1\leq e^{-\tau/t}t^{-1}$ using derivative and second derivative test, but no success. I thanks any help.
 A: HINT
So you really want to show
$$
t \le e^{\tau - \tau/t} = \left(e^\tau\right)^{1-1/t},
$$
Take logarithms, and your inequality is equivalent to showing
$$
\begin{split}
\ln t &\le (1-1/t)\ln \left(e^\tau\right) \\
\ln t &\le (1-1/t)\tau \\
t \ln t &\le (t-1)\tau \\
\tau &\le \frac{t\ln t}{t-1}, \quad \forall\ 0 < \tau < t < 1.
\end{split}
$$
where the last inequality sign flips dividing by $t-1$ since $t-1<0$. Thus, the problem is now equivalent to showing
$$
\frac{\ln t}{t-1} \le 1 \iff \ln t \ge t - 1
$$
A: Let $x={\tau\over t}.$ Then $0<x<1.$ The inequality takes the form
$$e^{-xt}\le e^{-x}t^{-1}$$
$$te^{x(1-t)}\le 1$$
We have
$$te^{x(1-t)}\le te^{1-t}$$ Let $y=1-t.$ It suffices to show that
$$te^{1-t}=(1-y)e^y\le 1,\qquad 0\le y<1$$ We have
$$e^y=\sum_{n=0}^\infty {y^n\over n!}\le \sum_{n=0}^\infty y^n=(1-y)^{-1}$$ So the result follows.
A: Both sides of the inequality are in the form $e^{-\tau/x}x^{-1}$, with $x=1$ on the LHS. Differentiating that, for $\tau < x < 1$,
$$\begin{align*}
f(x) &= e^{-\tau/x}x^{-1}\\
f'(x) &= e^{-\tau/x}\left(-x^{-2}\right) + x^{-1}\left(e^{-\tau/x}\tau x^{-2}\right)\\
&= e^{-\tau/x} x^{-3}\left(\tau-x\right)\\
&\le 0
\end{align*}$$
By mean value theorem, for some $c$ that is $t < c < 1$,
$$\begin{align*}
\frac{f(1)-f(t)}{1-t} &= f'(c)\\
f(1) - f(t) &= f'(c)(1-t)\\
&\le 0\\
e^{-\tau}&\le  e^{-\tau/t}t^{-1}
\end{align*}$$
