Question: Prove the inequality $(e^x -1)\ln(1+x)>x^2, x>0$

My steps so far:

$(e^x-1)\ln(1+x) > \ln(e^{x^2})$

$(e^x-1)(1+x) > e^{x^2}$

$e^x+xe^x-x-1>e^{x^2}$

Are the steps correct, and if yes what should I do next? Any hints will be greatly appreciated.

Answer: 
Using $e^x-1 \geq x+x^2/2$, you'll find
$$(e^x-1)\log(1+x)-x^2 \geq x(1+x/2)\log(1+x)-x^2$$
and since $x> 0$, it suffices to check $$\log(1+x) - \frac{x}{1+x/2} = \log(1+x) - \left(2 - \frac{4}{x+2}\right) \geq 0\,.$$
Equality holds at $x=0$, while you have (deriving with respect to $x$)
$$\frac{1}{1+x} - \frac{4}{(x+2)^2} = \frac{x^2}{(1+x)(x+2)^2} > 0 $$
for $x>0$.
Your steps seem very questionable.

Your first step is:

\begin{equation}
(e^x – 1) \ln(1+x) > \ln(e^{x^2})
\end{equation}

This is equivalent to the statement you are trying to prove, so you do not yet know if is true. You can use this as the new “target” for your proof, but you must be very clear that this is what you are doing. Instead, you simply wrote out the statement without explanation, implying that it is in fact true.

Your next step is:

\begin{equation}
(e^x – 1)(1 + x) > e^{x^2}
\end{equation}

It looks like you have taken the exponential of both sides. But, on the left hand side, you *should* get this instead:

\begin{align}
e^{(e^x – 1) \ln(1 + x)} &= \left (e^{\ln(1 + x)} \right)^{(e^x – 1)} \\
&= (1 + x)^{(e^x - 1)}.
\end{align}