Prove that $e^x>x(x+1)$

Question

Let $x>0$,prove that $$e^x>x(x+1)$$or $$x>\ln{x}+\ln{(x+1)}$$

we can use this Taylor some first four term, $$e^x>1+x+\dfrac{1}{2}x^2+\dfrac{1}{6}x^3$$ prove it

But this inequality it seem very nice,maybe there exsit simple methods or Amazing way?

Isn't the method by Taylor's theorem simple? – Yuxiao Xie Apr 21 '17 at 13:43 — Yuxiao Xie, Apr 21 '17 at 13:43

score 4 · Answer 1 · answered Apr 21 '17 at 13:41

4

We need to prove$$e^x>1+x+\dfrac{1}{2}x^2+\dfrac{1}{6}x^3\ge x^2+x$$which is$$x^3+3x^2+6x+6\ge 6x^2+6x$$which is$$x^3-3x^2+6=(x-2)^2(x+1)+2>0$$which is obvious for positive $x$.

answered Apr 21 '17 at 13:41

didgogns

3,597

have other methods? – math110 Apr 21 '17 at 13:45
This is the simplest method I can think of... – didgogns Apr 21 '17 at 13:46

score 3 · Answer 2 · answered Apr 21 '17 at 13:46

3

Yes. Let $$f(x)=e^x-x^2-x$$

And $f(0)=1$

Now, $f^{'}(x)=e^x-(2x+1)$ and $f^{"}(x)=e^x-2$
We get local minima in $x \in (1,2) $ where $f(x)>0$.

Now Since at minima, $f(x)>0$

Thus we can conclude $$e^x-(x^2+x)>0$$ $$e^x>(x^2+x)$$

answered Apr 21 '17 at 13:46

The Dead Legend

4,727

why $f(x)>0$ when $x\in (1,2)?$ – math110 Apr 21 '17 at 13:53
@function you may check for yourself? Or use Wolfram? – The Dead Legend Apr 21 '17 at 13:54
No,I hope can use hand to prove it,without wolfam – math110 Apr 21 '17 at 13:55
Let $0<x_0$ with $f'(x_0)=0$. Then $ e^{x_0}=1+2x_0\implies$ $f(x_0)=(1+2x_0)-x_0-x_0^2=1+x_0-x_0^2.$... The quadratic $g(x)=1+x-x^2$ is positive for $0\leq x<(1+\sqrt 5;)/2=1.618....,$ and we can calculate that $f'(1.6)>0.$ So $x_0<1.6$ and $f(x_0)=g(x_0)>0.$ – DanielWainfleet Apr 21 '17 at 15:29

Prove that $e^x>x(x+1)$

2 Answers2