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.

So my solution is a bit of "brute force" solution, not very elegant and nice. But that was how I was tackling algebra problems when I can't find better solution. First let's assume:
$$ l(x) = \ln\left(\frac{e - e^x}{1-x}\right) $$
$$ l(1) = \lim_{x \to 1} l(x) = \ln \lim_{x \to 1} \frac{e - e^x}{1-x} $$
$$ l(1) = \ln \lim_{x \to 1} \frac{-e^x}{-1} = \ln(e) = 1$$
$$ f(x) = 3l(x)^2 $$
$$ g(x) = x^2+x+1 $$
$$ h(x) = f(x) - g(x) $$
Idea:
Show that $h'(x) \ge 0$ for $x \gt 0$ and that $h(1)=0$. This implies that $h(x)$ is monotonically increasing. Using the fact that $h(1)=0$ it implies that $f(x) \lt g(x)$ for $x \lt 1$ and $f(x) \gt g(x)$ for $x \gt 1$. 
Note that $l(x) \gt 0$ and $x^2+x+1 > 0$ for $x \gt 0$, which means that the inequality for $f$ and $g$ implies the inequality for the original functions (e.g. we just square the initial functions).
Proof:
$$l'(x) = \frac{1}{1-x} - \frac{1}{e^{1-x}-1}$$
$$l'(1) = \lim_{x \to 1} \frac{e^{1-x} - (1-x)}{(1-x)(e^{1-x}-1)} = \lim_{x \to 1} \frac{1 -e^{1-x}}{(1-e^{1-x}) + (x-1)e^{1-x}}$$
$$l'(1) = \lim_{x \to 1} \frac{e^{1-x}}{e^{1-x} + e^{1-x} + (1-x)e^{1-x}} = \lim_{x \to 1} \frac{1}{3 - x} = \frac{1}{2}$$
$$f'(x) = 6 l(x) l'(x)$$
$$g'(x) = 2x+1 $$
$$h'(x) = f'(x) - g'(x)$$
We want to show that $h'(x) \ge 0$. A sufficient condition for this is that $h''(x) = 0$ has a unique solution $x^*$ for $ x \gt 0$ and that it is a monotonically increasing function ($h''(x)$). The condition implies that $h'(x)$ has a unique minimum and that is $x^*$. Furthermore, we will show that $h'(x^*) = 0$.
$$l''(x) = \frac{1}{(1-x)^2} - \frac{e^{1-x}}{(e^{1-x}-1)^2}$$
$$l''(1) = \lim_{x \to 1} \frac{(e^{1-x}-1)^2 - (1-x)^2 e^{1-x}}{(1-x)^2e^{1-x}-1)^2} = \frac{1}{12}$$
$$f''(x) = 6l(x)l''(x) + 6 l'(x)^2$$
$$g''(x) = 2$$
$$h''(x) = 6l(x)l''(x) + 6 l'(x)^2 - 2 $$
$$ h''(x) = 0 \iff l(x)l''(x) + l'(x)^2 = \frac{1}{3} $$
Now to proof this first note that $h''(1) = \frac{1}{3}$, thus our candidate $x^*=1$. Also, as expected $h'(1)=0$. To conclude the proof we need to show that $h''(x)$ is monotonically increasing for $x \gt 0$.
For now I need to prove it, but if you plot the function it looks like it is. My guess is you will need to proof in fact that $h'''(0) \gt 0$. 
