What's the idea behind this inequality? $e^{\frac{\ln x}{x-1}}\geq\frac{e}{3}(1+\frac{1}{\sqrt{x}}+\frac{1}{x})$
This inequality's image (left hand substracted right hand)

which shows a perfect approximation between (1,1.5), where it nearly keeps unchanged.
I wonder how this inequality was created, did it use some series expanding or some famous inequalities?
 A: With substitution $x = \frac{1}{y^2}$, the inequality is written as
$$\mathrm{e}^{\frac{-2y^2\ln y}{1-y^2}} 
\ge \frac{\mathrm{e}}{3}( 1 + y + y^2).$$
Denote $f(y) = \mathrm{e}^{\frac{-2y^2\ln y}{1-y^2}}$.
Then the inequality above is written as $f(y) \ge f(1) + f'(1)(y-1) + \frac{1}{2}f''(1)(y-1)^2$.
The RHS is the second order Taylor approximation of LHS.
This approximation is nice around $y=1$ because $f'''(1) = 0$ and $f'''(y)$ is small near $y=1$.
A: This is not an answer.
Firt of all, Welcome to the site !.
In the area of function approximation, a lot of things can be done. Let me propose one I just made for you
$$e^{\frac{\log( x)}{x-1}}=e\frac{1+\frac{87}{100} (x-1)+\frac{37}{240}(x-1)^2}{1+\frac{137 }{100}(x-1)+\frac{457 }{1200}(x-1)^2}$$
The inequality does not hold but it is almost OK.
A: Let
$$f(x):=e^{\log x/(x-1)},\qquad g(x):={e\over3}\left(1+{1\over\sqrt{x}}+{1\over x}\right)\ .$$
Then
$$\log\bigl(f(1+t)\bigr)={\log(1+t)\over t}=1-{t\over2}+{t^2\over3}-{t^3\over4}+{t^4\over5}-{t^5\over6}+?\,t^6\ .$$
Some calculation gives
$$\log\bigl(g(1+t)\bigr)=1-{t\over2}+{t^2\over3}-{t^3\over4}+{115t^4\over576}-{239t^5\over1440}+?\,t^6$$
It follows that
$$\log{f(1+t)\over g(1+t)}=\log\bigl(f(1+t)\bigr)-\log\bigl( g(1+t)\bigr)={1\over2880}t^4+?\,t^5\qquad(t\to0)\ .\tag{1}$$
This shows that ${f(x)\over g(x)}\approx1$ and $\geq1$ with high precision when $x\approx1$.
It seems that the function $g$ has been set up in such a way that we get a result in the sort of $(1)$. Other data resulting in a similar result would have been possible.
