<s>[INST] Prove this stronger inequality with $\frac{e^x}{x+1}-\frac{x-1}{\ln{x}}-\left(\frac{e-2}{2}\right)>0$ 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.
 [/INST] 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.
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