Prove this stronger inequality with $\frac{e^x}{x+1}-\frac{x-1}{\ln{x}}-\left(\frac{e-2}{2}\right)>0$

(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 $x\gt 1$.

Set $$f(x)=\frac{e^x}{x+1}-\frac{ex}{4},\,\,g(x)=\frac{ex}{4}-\frac{x-1}{\ln x}.$$ Then $$f'(x)=\frac{xe^x}{(x+1)^2}-\frac{e}{4}=e\cdot\frac{4xe^{x-1}-(x+1)^2}{4(x+1)^2},$$ and since $e^{x-1}>x$ for all $x>1$, we obtain $$4xe^{x-1}-(x+1)^2>4x^2-(x+1)^2>0,$$ since $x>1$. This shows that $f$ is increasing. Moreover, $$g'(x)=\frac{x-1-x\ln x}{x\ln^2x}+\frac{e}{4}=\frac{x-1-x\ln x}{x\ln^2x}+\frac{1}{2}+\frac{e-2}{4}.$$ We will show that $g$ is also increasing for $x>1$; to do this, it is enough to show that $$x-1-x\ln x+\frac{1}{2}x\ln^2x>0,$$ or, equivalently, $$\ln^2x-2\ln x+2-\frac{2}{x}>0,$$ for $x>1$.

Set $h(x)=\ln^2x-2\ln x+2-\frac{2}{x}$, and suppose that $h$ has a root $x_0$, for some $x_0>1$. Since $h(1)=0$, there exists a root of $h'(x)=0$ in $(1,\infty)$. But, $$h'(x)=2\frac{\ln x}{x}-\frac{2}{x}+\frac{2}{x^2}=\frac{2}{x}\left(\ln x-1+\frac{1}{x}\right),$$ and the last function does not have roots greater than $1$, which is a contradiction. This shows that $h$ has a fixed sign on $(1,\infty)$, and since $h$ goes to $\infty$ as $x\to\infty$, we obtain that $h(x)>0$ for $x>1$.

So, we obtain that $g$ is increasing as well. Therefore, $$\frac{e^x}{x+1}-\frac{x-1}{\ln x}=f(x)+g(x)$$ is increasing as well, which shows the inequality.