Ineqaulity about a convex function. $(e^x-1)\cdot \ln(x+1) >x^2$ $(x \gt 0)$? How can I solve this inequality $(e^x-1)\cdot \ln(x+1) >x^2$ ($x > 0$)?

Since $e^x-1 \gt0,$ we will prove the following :
$$\ln(x+1)-\frac{x^2}{e^x-1}\gt0.$$
Letting $f(x)$ be the left hand side, we have
$$f^\prime(x)=\frac{1}{x+1}-\frac{2x(e^x-1)-x^2\cdot e^x}{(e^x-1)^2}=\frac{g(x)}{(x+1)(e^x-1)^2}$$
where $g(x)=e^{2x}+(x^3-x^2-2x-2)e^x+2x^2+2x+1.$
Now 
$$g^\prime(x)=2e^{2x}+(3x^2-2x-2)e^x+(x^3-x^2-2x-2)e^x+4x+2$$
$$=2e^{2x}+(x^3+2x^2-4x-4)e^x+4x+2$$
$$\gt 2(x+1)^2+(x^3+2x^2-4x-4)(x+1)+4x+2\gt0.$$
Here, I used $e^x\gt x+1$. You will know the last inequality is true with some calculation.
Hence, we now know $g(x)$ is a strictly monotone increasing function. Also, we have $g(0)=0$. Hence, we know $g(x)\gt0.$
Hence, we now know that $f^\prime(x)\gt0.$ So, we know $f(x)$ is a strictly monotone increasing function. Also, we have 
$$\lim_{x\to 0+}f(x)=\lim_{x\to 0+}\frac{x^2}{e^x-1}=\lim_{x\to 0+}\frac{2x}{e^x}=0.$$
So, we now know $f(x)\gt0.$ Q.E.D.
