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This is easy to prove for negative $x$ but what about positive $x$? Should I use MVT?

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    $\begingroup$ You could note that the function $f:\mathbb{R}\to\mathbb{R}, f(x)=e^x$ is convex, and that $g:\mathbb{R}\to\mathbb{R}, g(x)=ex$ is tangent to $f$ at $x=1$. $\endgroup$ – Uncountable May 2 '15 at 18:40
  • $\begingroup$ Taylor series would work here... $\endgroup$ – Zach466920 May 2 '15 at 18:44
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    $\begingroup$ It is better to replace $x$ by $x + 1$ to get $e^{x} \geq 1 + x$. $\endgroup$ – Paramanand Singh May 3 '15 at 11:02
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You should prove that:

$$0 \leq e^{x-1}-x=f(x)$$

Let's calculate derivative of $f(x)=e^{x-1}-x$, it's:

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

Note that $f'(x)<0$ for $x<1$ and $f'(x)>0$ for $x>1$. So $f(x)$ reaches the smallest value at $x=1$, but we can calculate that $f(1)=0$, so

$$f(x) \geq f(1)=0$$

for all $x \in \mathbb{R}$.

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This inequality is equivalent to $e^{x-1}\ge x$, which, after substitution $y:=x-1$ is equivalent to $$e^y\ge 1+y.$$

You can find many proofs for this one:

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Define

$$f(x):=ex-e^x\implies f'(x)=e-e^x=e(1-e^{x-1})=0\iff x=1$$

But also

$$f''(1)=-e^1<0$$

so that at $\;x=1\;$ we have a maximum point... End the argument now.

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Use the convexity of $\exp$ at the point 1 to get:
$$ \exp x \ge \exp' 1 \times (x - 1) + \exp 1 = ex $$

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Note the tangent to $y=e^x$ at $x=1$ is $$ y-e = e(x-1), $$ and the Taylor series for $e^x$ about $x=1$, with remainder, is $$ e^x -ex = \int_1^{x} (x-t)e^{t} \, dt, $$ and the integrand is clearly positive for $x>1$. For $x<1$, you can rearrange integrand and limits to show the integral is still positive.

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