TITLE: Prove this stronger inequality with $\frac{e^x}{x+1}-\frac{x-1}{\ln{x}}-\left(\frac{e-2}{2}\right)>0$
QUESTION [6 upvotes]: Let $x>1$ show that
$$\dfrac{e^x}{x+1}-\dfrac{x-1}{\ln{x}}-\left(\dfrac{e-2}{2}\right)>0$$
It seem this inequality can use derivatives solve it,But it is ugly.can you  help？
$$\lim_{x\to 1}\left(\dfrac{e^x}{x+1}-\dfrac{x-1}{\ln{x}}-\left(\dfrac{e-2}{2}\right)\right)=0$$
and $f(x)=\dfrac{e^x}{x+1},g(x)=\dfrac{x-1}{\ln{x}}$,But $$f'(x)=\dfrac{xe^x}{(x+1)^2}>0, g'(x)=\dfrac{\ln{x}-\dfrac{x-1}{x}}{\ln^2{x}}>0$$
Unfortunately I don't know any nice expression for this 1-th derivative which could help. I'll be grateful for all useful suggestions.

REPLY [3 votes]: (I have to admit that the method in this answer is "ugly".)
Multiplying the both sides by $2(x+1)\ln x\gt 0$ gives
$$2e^x\ln x-2(x-1)(x+1)-(e-2)(x+1)\ln x\gt 0\tag1$$
Let $f(x)$ be the LHS of $(1)$. Then,
$$f'(x)=2\left(e^x\ln x+\frac{e^x}{x}\right)-4x-(e-2)\left(\ln x+\frac{x+1}{x}\right)$$
$$g(x):=xf'(x)=2xe^x\ln x+2e^x-4x^2-(e-2)(x\ln x+x+1)$$
$$g'(x)=2e^x\ln x+2xe^x\ln x+4e^x-8x-(e-2)(\ln x+2)$$
$$g''(x)=2e^x\ln x+\frac{2e^x}{x}+2e^x\ln x+2x\left(e^x\ln x+\frac{e^x}{x}\right)+4e^x-8-\frac{e-2}{x}$$
$$h(x):=xg''(x)=4xe^x\ln x+2e^x+2x^2e^x\ln x+6xe^x-8x-(e-2)$$
$$h'(x)=4e^x\ln x+8xe^x\ln x+12e^x+2x^2e^x\ln x+8xe^x-8$$
$$h''(x)=12e^x\ln x+\frac{4e^x}{x}+12xe^x\ln x+28e^x+2x^2e^x\ln x+10xe^x$$
Now it's easy to see that $h''(x)\gt 0$ for $x\gt 1$ because each term is positive for $x\gt 1$.
Since we have
$$h'(1)=20e-8\gt 0,\quad h(1)=7e-6\gt 0,\quad g'(1)=2e-4\gt 0,\quad g(1)=f(1)=0$$ we can see that each of $h'(x),h(x),g'(x),g(x),f(x)$ is increasing for $x\gt 1$, and that $h'(x)\gt 0,h(x)\gt 0,g''(x)\gt 0,g'(x)\gt 0,g(x)\gt 0,f'(x)\gt 0,f(x)\gt 0$ for $x\gt 1$.