Prove that $e^x \le \dfrac{1}{1-x}, x\lt 1.$
I find that if we set $f(x)=e^x(1-x)$ then $f(0)=1 $ and $f'(x)<0, x\in(0,1]$ proving the inequality for $x\in[0,1]$ but I don't see how to prove it for $x\in(-\infty,0)$. The derivative $f'(x)\gt0 $ in that interval ... would $f(0)=0$ in addition to that be sufficient to prove it for $x\in(-\infty,0)$?
If that doesn't work, could we use something like mean value theorem or simliar to prove it!
$\endgroup$3 Answers
$\begingroup$For 0 < x < 1,
$$e^x = \sum_{k=0}^\infty\frac{x^k}{k!} < \sum_{k=0}^\infty x^k = \frac{1}{1-x}.$$
If $y > 0$ we have $e^y > 1+y$ and $e^{-y} < (1+y)^{-1}$. Set $x = -y < 0.$
Then with $x < 0$ we have
$$e^x < \frac{1}{1-x}$$
$\endgroup$ 4 $\begingroup$In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequality
$$e^z\ge 1+z \tag 1$$
for $z>-1$. Setting $z=-x$ in $(1)$ yields
$$e^{-x}\ge 1-x \tag 2$$
for $x<1$. Taking the reciprocal inequality of $(2)$, we obtain
$$e^x\le \frac{1}{1-x}$$
for $x<1$ as was to be shown!!
$\endgroup$ 0 $\begingroup$For $x<0$ only: We have $1-x>0$. So $e^x\leq 1/(1-x)\iff f(x)=e^x(1-x)\leq 1.$ We have $f'(x)=-x e^x>0$ so $f$ is strictly increasing. So $f(x)<\lim_{y\to 0^-}f(y)=1.$
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