Ineqaulity about a convex function. $(e^x-1)\cdot \ln(x+1) >x^2$ $(x \gt 0)$?

Question

How can I solve this inequality $(e^x-1)\cdot \ln(x+1) >x^2$ ($x > 0$)?

Answer 1

Set $f(x)=e^x-1$, and $\displaystyle g(x)=\frac{f(x)}{x}$. Then $f$ and $g$ are differentiable on $(0,\infty)$, and $$g'(x)=\frac{xf'(x)-f(x)}{x^2}.$$ But, $xf'(x)-f(x)>0$ for $x>0$: $$(xf'(x)-f(x))'=f'(x)+xf''(x)-f'(x)=xf''(x)>0\Rightarrow$$$$xf'(x)-f(x)>0\cdot f(0)-f(0)=0,$$ therefore $g'(x)>0$ and $g$ is increasing. Now, since $x>\ln(x+1)$ for $x>0$, you have that $$g(\ln(x+1))<g(x)\Rightarrow \frac{f(\ln(x+1))}{\ln(x+1)}<\frac{f(x)}{x}\Rightarrow \frac{x}{\ln(x+1)}<\frac{e^x-1}{x},$$ which is the inequality you want.

Answer 2

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.

Answer 3

The hint.

Let in $\Delta ABC$ the angle $ABC$ be obtuse and $BD$ be an altitude of the triangle.

Thus, $AD\cdot CD>BD^2$.