Prove inequality with $e^x$ and $\ln$ on the same side The problem is to prove the following inequality:
$$ (e^x - 1) \ln(1+x) > x^2 , \quad\text{ for } x >0 $$
Let me introduce notation $f(x) > g(x)$.
At $x=0$ both sides are equal to $0$. So, to prove the inequality one could prove it not for original functions, but for their derivatives. Unfortunately, derivatives of the functions doesn't seem to simplify the situation at all.
I've also tried to substitute some parts of the original inequality with the new variable like this:


*

*$ z = e^x $;

*$ z = \ln(1+x) $.


Finally, I've tried to separate $e^x$ and $\ln(1+x)$ (which is the major obstacle for good differentiation) by dividing both sides of the inequality by one of these functions.
None of the above helped me to move significantly towards proof.
I'm kindly asking for your help. Both hints and complete solutions are very appreciated.
 A: With $f(x)=(\operatorname e^x-1)\ln(1+x)$ and $g(x)=x^2$ it is easy to show that $f(0)=g(0)$ and $f'(0)=g'(0)$ since all of these are equal to zero. Digging deeper we find that $f''(0)=g''(0)=2$, and finally we reach $f'''(0)=g'''(0)=0$. Now note that
$$
f'''(x)=(\operatorname e^x-1)\cdot\frac{2}{(1+x)^3}+\underbrace{3\operatorname e^x\cdot\frac{-1}{(1+x)^2}+3\operatorname e^x\cdot\frac1{1+x}}_{\text{greater than zero}}+\operatorname e^x\ln(1+x)>0
$$
for $x>0$, since we see that the negative contribution from the second term is clearly compensated by the positive contribution from the third term. The first and last terms above are clearly positive for $x>0$.
So working backwards we have $f'''(x)>g'''(x)$ implies $f''(x)>g''(x)$ implies $f'(x)>g'(x)$ implies $f(x)>g(x)$ for $x>0$.

Note that the principle for higher order derivatives of a product $p(x)q(x)$ is given as
$$
\begin{array}{ccc}
(p(x)q(x))'&=&p'(x)q(x)+p(x)q'(x)\\
(p(x)q(x))''&=&p''(x)q(x)+2p'(x)q'(x)+p(x)q''(x)\\
(p(x)q(x))'''&=&p'''(x)q(x)+3p''(x)q'(x)+3p'(x)q''(x)+p(x)q'''(x)\\
&\text{etc.}
\end{array}
$$
where the coefficients $1\ 1$ in the first row, $1\ 2\ 1$ in the second, and $1\ 3\ 3\ 1$ in the third row are the renowned binomial coefficients. So with $p(x)=(\operatorname e^x-1)$ and $f(x)=\ln(1+x)$ this enables us to compute derivatives of $f(x)=p(x)q(x)$ efficiently.
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: $e^x-1\geq x+\dfrac{x^2}{2}+\dfrac{x^3}{6}$ and $\ln(1+x)\geq\dfrac{x}{1+x}$ for $x>0$. 
