Find the maximum integer $n$ such that $e^{x-1}\ln x+\frac{e^x+2x-1}{x^2} > 2\sqrt{n}-1$ 
If $$e^{x-1}\ln x+\frac{e^x+2x-1}{x^2} > 2\sqrt{n}-1 \quad (n \in \mathbb{N}^+),$$
  find the maximum of $n$.

I can prove that the inequality holds when $n=4$.
My proof:
When $n=4$:
$$e^{x-1}\ln x+\frac{e^x+2x-1}{x^2} > 3\quad(n \in \Bbb N^+)$$
Case I. $0<x<1.$
$$e^{x-1}\ln x>(x+1)\left(1-\frac{1}{x}\right)=x-\frac{1}{x}$$
Hence
$$e^{x-1}\ln x+\frac{e^x+2x-1}{x^2} > 3\quad(n \in \Bbb N^+)$$
$$\iff e^x>1+x+\frac{x^2}{2}$$
Case  II. $x\geq 1.$
$$e^{x-1}\ln x>x\left(1-\frac{1}{x}\right)$$
Hence
$$e^{x-1}\ln x+\frac{e^x+2x-1}{x^2} > 3\quad(n \in \Bbb N^+)$$
$$\iff e^x>1+x+\frac{x^2}{2}+\frac{x^3}{6}$$
Q.E.D.
But I am not sure whether the inequality holds when $n=5$.
Any ideas?
 A: We will prove that $\max n=4$.

Let $$f(x)=1+e^{x-1}\ln x+\frac{e^x+2x-1}{x^2}\tag1.$$ We wish to determine $\max n$ such that $$\min f(x)=2\sqrt{\max n}\implies \max n=\frac14[\min f(x)]^2\tag2.$$ Then $$f'(x)=\frac{e^x\ln x}e+\frac{e^x}{ex}+\frac{(e^x+2)x^2-2x(e^x+2x-1)}{x^4}=0\tag3$$ for critical points, which can be rearranged to give $$x^2e^x(x\ln x+1)+e((x-2)e^x-2x+2)=0\tag4.$$ Denote the LHS of $(3)$ as $g(x)$. Then \begin{align}g'(x)&=x(x+2)e^x(x\ln x+1)+x^2e^x(\ln x+1)+e((x-1)e^x-2)\\&=xe^x((x^2+3x)\ln x+2x+2)+e((x-1)e^x-2)\tag5\end{align} and $g'(x)$ is monotonically increasing for all $x>0$ since $$g''(x)=(x^3+6x^2+6x)\ln x+(3x^2+(9+e)x+2)\tag6$$ which is clearly positive. Since $g'(0.5)<0$ and $g'(1)>0$, there exists a unique real number $k\in(0.5,1)$ such that $g(k)$ is a critical point. It can be shown that $g(k)$ is a minimum value since $g(x)<0$ for $0<x<1$.
Now $g(x)$ is monotonically increasing for all $x>k$ and numerical experimentation on a simple scientific calculator reveals that $g(1.3109)<0$ and $g(1.3110)>0$. Between these two values, $$f(1.3109)\approx4.471809644,\,\quad f(1.3110)\approx4.471809635\tag7.$$ From $(2)$, since $n\in\Bbb N^+$, $$\boxed{\max n<\frac14\cdot4.471809644^2=4.999270\cdots<5\implies \max n=4.}\tag8$$ The bound is extremely tight.
