how to proof $(e^x - 1) \log(1 + x) \gt x^2$ when $x\gt 1$ [duplicate]
I was given a task to prove that inequality is true for x>0: $(e^x-1)\ln(1+x) > x^2$. I've tried to use derivatives to show that the $f(x) = (e^x-1)\ln(1+x)-x^2$ is greater than zero, but has never succeeded.
Any help will be appreciated.
Let $f(x)=\frac{e^x-1}{x}$. Then:
- $f(x)>1$ for $x>0$ (stems from $e^x\ge1+x$, standard)
- $f(x)$ is increasing for all $x$ (also boils down to $e^x\ge1+x$)
(1) tells us that $e^y-1>y\implies f(e^y-1)>f(y)$ by (2). Let $y=f^{-1}(x)=\log(1+x)$, whence
$$f(x)>f(\log(1+x)) \implies \frac{e^x-1}{x} > \frac{x}{\log(1+x)}$$
The desired inequality follows.
The exponential makes your approach somewhat impractical: get rid of it. You have $e^x = 1+x+\frac{x^2}{2}+\sum_{n=3}^\infty \frac{x^n}{n!}$, so for $x > 0$ it holds that $e^x - 1 > x+\frac{x^2}{2}$.
Then it suffices to show that for any $x> 0$, $$ \left(x+\frac{x^2}{2}\right)\ln(1+x) > x^2 $$ or equivalently $$ \left(1+\frac{x}{2}\right)\ln(1+x) > x. $$ Define $f$ on $[0,\infty)$ by $f(x) = \left(1+\frac{x}{2}\right)\ln(1+x) - x$. It is $C^\infty$, with $f(0)=0$, and derivative $f^\prime(x) = \frac{(x+1)\ln(x+1)-x}{2(x+1)}$. From there, we get $f^\prime(0) = 0$ and $f^\prime(x) > 0$ for all $x>0$, so $f$ is increasing on $[0,\infty)$ and the inequality holds.
Let $A(x,e^x-1)$, $B(x,-\ln(1+x)$ and $O(0,0)$.
Now easy to see that $\measuredangle AOB>90^{\circ}$ and we are done!