Prove that inequality is true for $x>0$: $(e^x-1)\ln(1+x) > x^2$ 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.
 A: 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.
A: 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.
A: 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!
A: I thought I would
get the power series
and see if
all the coefficients
are non-negative.
Turns out that that is not true.
So I'll show what I did
anyway.
So this is another of my
failed attempts.
$\begin{array}\\
(e^x-1)\ln(1+x)
&=\sum_{i=1}^{\infty} \dfrac{x^i}{i!}\sum_{j=1}^{\infty} \dfrac{(-1)^{j-1}x^j}{j}\\
&=x^2\sum_{i=1}^{\infty} \dfrac{x^{i-1}}{i!}\sum_{j=1}^{\infty} \dfrac{(-1)^{j-1}x^{j-1}}{j}\\
&=x^2\sum_{i=0}^{\infty} \dfrac{x^{i}}{(i+1)!}\sum_{j=0}^{\infty} \dfrac{(-1)^{j}x^{j}}{j+1}\\
&=x^2\sum_{n=0}^{\infty}\sum_{i=0}^{n} \dfrac{x^{i}}{(i+1)!} \dfrac{(-1)^{n-i}x^{n-i}}{n-i+1}\\
&=x^2\sum_{n=0}^{\infty}x^n\sum_{i=0}^{n} \dfrac{(-1)^{n-i}}{(i+1)!(n-i+1)}\\
&=x^2\sum_{n=0}^{\infty}x^n\sum_{i=0}^{n} \dfrac{(-1)^{i}}{(n-i+1)!(i+1)}\\
\end{array}
$
We are done if
$c_n
=\sum_{i=0}^{n}\dfrac{(-1)^{i}}{(n-i+1)!(i+1)}
\ge 0$.
$c_0 = 1$.
$c_1 = \frac1{2}-\frac1{2}
=0
$.
$c_2 = \frac1{6}-\frac1{4}+\frac1{3}
=\frac1{4}
$.
$c_3 = \frac1{24}-\frac1{12}+\frac1{6}-\frac1{4}
=-\frac18
$.
So it's not true.
