1

i just tried to solve this question, which is a small part of a bigger one.

Prove for all all $x \in \mathbb{R}: \exp(x-1) \geq x$

My first attempt was to simplify:

$$ e^{x-1} \geq x $$ $$ \Leftrightarrow \ln(e^{x-1}) \geq \ln(x) $$ $$ \Leftrightarrow x-1 \geq \ln(x) $$ $$ \Leftrightarrow x \geq \ln(x)+1 $$ $$ \Leftrightarrow e^x \geq e^{\ln(x)}+e^1 $$ $$ \Leftrightarrow e^x \geq x+e $$

My idea is that $e$ power $x$ is of course greater equal $x+e$. Is this correct and if yes is it enough?

Michael Hardy
  • 1
hukachaka
  • 227
  • Careful: when you try to run this backwards you have $\ln x$ which only makes sense for positive $x$s, and you may want this for all $x$s. – Randall Oct 26 '17 at 18:49
  • ok yes. you got a point :-/ you have an idea how to solve? – hukachaka Oct 26 '17 at 18:53
  • 2
    You made a mistake at the point where you asserted the "equivalent" inequality $e^x\le e^{\ln x}+e^1$. That inequality is not true. For example, if $x=1$, then $e^{\ln 1}=e^0=1$ so we have $e^1\lt1+e^1$, not $\ge$. – Barry Cipra Oct 26 '17 at 19:09
  • 3
    Also notice that $e^{\ln(x)+1}=e^{\ln x}\cdot e=x\cdot e\neq x+e$. – kingW3 Oct 26 '17 at 19:09
  • To further emphasize the two previous comments, note that $e^x\geq x+e$ implies $1\geq e$ which is pretty unlikely. – Surb Oct 26 '17 at 19:10
  • ok this was a huge mistake i do understand now :) - i have really some maths issues – hukachaka Oct 26 '17 at 19:19
  • Also: https://math.stackexchange.com/questions/630481/prove-expx-geq-1x-forall-x-in-mathbbr (and probably some more). – Martin R Oct 26 '17 at 21:41

2 Answers2

9

Let $f(x)=e^{x-1}-x.$

Thus, $f'(x)=e^{x-1}-1$, which says that $x_{\min}=1$.

Id est, $$f(x)\geq f(1)=0$$ and we are done!

Michael Hardy
  • 1
Michael Rozenberg
  • 194,933
0

the answer Michael gave is a good answer but if you want different way, $$e^{x-1}=x\implies x=1$$ now because both $x$ and $e^{x-1}$ are continuous you just need to check one value of $x$ that is above $1$ and one that is less. for example:$$\text{for $x<1,\,x=0$},\,e^{0-1}={1\over e}\gt0\\\text{for $x>1,\,x=2$},\,e^{2-1}=e\gt2$$ therefore $$e^{x-1}\ge1$$

ℋolo
  • 10,006