Prove that the logarithmic mean is less than the power mean. Prove that the logarithmic mean is less than the power mean.
$$L(a,b)=\frac{a-b}{\ln(a)-\ln(b)} < M_p(a,b) = \left(\frac{a^p+b^p}{2}\right)^{\frac{1}{p}}$$ such that $$p\geq \frac{1}{3}$$ That is the $\frac{1}{p}$ root of the power mean.
 A: By the inequality of Power Means, it is sufficient to prove this for $p = \frac13$.  Also WLOG we can assume $a > b > 0 \implies x = \frac{a}b > 1$.  So the inequality we are left to show is, for $x > 1$:
$$\frac{x-1}{\log x} < \left(\frac{x^{1/3}+1}2 \right)^3$$ 
Simplifying using $x = t^3$, this is equivalent to showing for $t > 1$
$$\log t > \frac{8(t^3-1)}{3(t+1)^3}$$
At $t=1, LHS = 0 = RHS$, so it is sufficient to show that LHS increases faster than RHS, or 
$$\frac1t > \frac{8(1+t^2)}{(1+t)^4} \iff 8t(1+t^2) < (1+t)^4 \iff (t-1)^4 > 0$$
A: Consider the function $e:\mathbb{R}\cup\{0\} \to\mathbb{R}$; 
$e(s)=\dfrac{x^s-y^s}{s}$, when $s\neq 0$, and $e(s)=\ln(x)-\ln(y)$, when $s=0$ (where, $x,y>0$).
Note that $e$ is continuous on $\mathbb{R}\cup\{0\}$. 
Further, $e(s)=\int_y^x v^{s-1}\,dv$, 
Since, for arbitrary $a,b,s,t\in \mathbb{R}$, 
$a^2e(s)+2abe(\frac{s+t}{2})+b^2e(t)=\int_y^x a^2v^{s-1}+2abv^{\frac{s+t}{2}-1} +b^2v^{t-1}\,dv=\int_y^x(av^{\frac{s-1}{2}}+bv^{\frac{t-1}{2}})^2\,dv\ge0$,
it follows from the negative condition of discriminant that, $e(s)e(t)\ge (e(\frac{s+t}{2}))^2$,
i.e., $\log(e(s))$ is a convex function on $\mathbb{R}\cup\{0\}$, that is for arbitrary non negative values $r,s,t$ we have 
$(t-s)\ln e(r)+(r-t)\ln e(s)+(s-r)\ln e(t)\ge0$ .
Taking anti-log, $\large(\frac{e(s)}{e(r)})^{\frac{1}{s-r}}\le (\frac{e(t)}{e(r)})^{\frac{1}{t-r}}$, for $t>s$
Therefore, $E(r,s):=(\frac{e(s)}{e(r)})^{\frac{1}{s-r}}=(\frac{e(r)}{e(s)})^{\frac{1}{r-s}}=E(s,r)$ is increasing in $r$ and $s$, 
I.e. $E(2s,s)=\large\left(\frac{x^s+y^s}{2}\right)^{\frac{1}{s}}$ and $E(s,0) = \frac{x^s-y^s}{s(\ln x - \ln y)}$
$E(2s,s)\ge E(s,s) \ge E(s,0)\ge E(1,0)$, if $s>1$.
It remains to verify that $E(2/3,1/3)\ge E(1,0)$
Follow Macavity's proof :)
EDIT: (proof by F. Burke) From a geometric point of view the final inequality can also be derived from the Simpson's $3/8$ rule:
$\displaystyle \int_c^d f(x)\,dx = \left(\dfrac{f(c)+3f(\frac{2c+d}{3})+3f(\frac{c+2d}{3})+f(d)}{8}\right)(d-c) - \dfrac{(d-c)^5}{6480}f^{(4)}(\theta)$, where $\theta \in (c,d)$.
For the function $f(x) = e^x$, with limits $c =\ln a$ and $d = \ln b$, the rule suggests,
$$\displaystyle \int_{\ln a}^{\ln b} e^x\,dx \le \left(\dfrac{e^{\ln a}+3e^{\frac{2\ln a + \ln b}{3}}+3e^{\frac{\ln a + 2\ln b}{3}}+e^{\ln b}}{8}\right)(\ln b- \ln a)$$
i.e., $\displaystyle (b-a) \le \left(\dfrac{a^{1/3}+b^{1/3}}{2}\right)^3(\ln b - \ln a)$.
A: Here is my proof. I feel like I made a huge leap at the end. I was not sure how to embed my LaTex code, it would not work. So I took screenshots. The last two lines, I have a gut feeling that I am missing a key step that links the two.



