Prove inequality $\ln \left( \frac{e-e^x}{1-x} \right) \leq \sqrt{\frac{1+x+x^2}{3}}$ for $0The first function could be called 'exponential mean' of $y$ and $x$:
$$f(y,x)=\ln \left( \frac{e^y-e^x}{y-x} \right)$$
We can obtain it by Cauchy mean value theorem.
What is interesting, it appears numerically that:

$$\ln \left( \frac{e-e^x}{1-x} \right) \leq \sqrt{\frac{1+x+x^2}{3}}, \qquad 0<x \leq 1$$
$$\ln \left( \frac{e-e^x}{1-x} \right) \geq \sqrt{\frac{1+x+x^2}{3}}, \qquad x \geq 1$$
How would you prove both of these inequalities?

Around $1$ both functions are very close. See the plot below:

This is related to my recent question.
The series expansion doesn't seem to be the best way.
The first function can be represented:
$$\ln \left( \frac{e-e^x}{1-x} \right)=1+\ln(1-e^{-(1-x)})-\ln(1-x)$$
But that's not any better.

A little demonstration (not a full solution, but I'll try to expand it later):
If we were allowed to use CAS, such as Mathematica (or spent a little time differentiating), we could just square both sides and expand the left side into series aroud $x=1$:
$$\ln^2 \left( \frac{e-e^x}{1-x} \right) =1+(x-1)+\frac{1}{3} (x-1)^2+\frac{1}{24} (x-1)^3+\frac{1}{960} (x-1)^4+O((x-1)^5)$$
Now we expand and simplify the first three terms:
$$\ln^2 \left( \frac{e-e^x}{1-x} \right) =\frac{1}{3}+\frac{x}{3}+\frac{x^2}{3}\color{blue}{+\frac{1}{24} (x-1)^3+\frac{1}{960} (x-1)^4}+O((x-1)^5)$$
Comparing with the square of the right-hand side we immediately see:
$$\ln^2 \left( \frac{e-e^x}{1-x} \right)- \frac{1+x+x^2}{3}=\frac{1}{24} (x-1)^3+\frac{1}{960} (x-1)^4+O((x-1)^5)$$
Thus, around $x=1$ we have:

$$\ln^2 \left( \frac{e-e^x}{1-x} \right)- \frac{1+x+x^2}{3}=-\frac{1}{24} (1-x)^3+\frac{1}{960} (1-x)^4-O((1-x)^5) \leq 0, $$ $$x \to 1^-$$
$$\ln^2 \left( \frac{e-e^x}{1-x} \right)- \frac{1+x+x^2}{3}=\frac{1}{24} (x-1)^3+\frac{1}{960} (x-1)^4+O((x-1)^5) \geq 0, $$ $$ x \to 1^+$$

