# 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.

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I'm currently working on the problem. I' wanted to attempt induction but this will not work I assume. My professor wanted to pull out an a from both equations, and add them together when p=1/3. But I was not sure how this would help. – Frumpy Mar 5 '14 at 2:00
Any hint would be helpful. I am stuck and I can not get past this. – Frumpy Mar 5 '14 at 2:37

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 $$\frac1t > \frac{8(1+t^2)}{(1+t)^4} \iff 8t(1+t^2) < (1+t)^4 \iff (t-1)^4 > 0$$

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Why is it enough to prove this for the p=1/3 case only? And how did you get to that first equation from the original? I'm going to attempt to figure this out in the mean time. – Frumpy Mar 5 '14 at 4:12
@Macavity +1 Nice :) – r9m Mar 5 '14 at 4:19
@Macavity I see that you pulled out a "b" from both sides and substituted x=a/b. I see that. But, I do not see who you simplified using x=t^3 to attain the second equation. Nor do I see why. I don't simply want to know how the proof is done but I also want to understand why; this is critical for me. Edit: I see how you manipulated to attain the second equation. I will continue to work with this. – Frumpy Mar 5 '14 at 4:27
If you prove for $p=\frac13$, then the Power Means Inequality (check for e.g. artofproblemsolving.com/Wiki/index.php/Power_Mean_Inequality) assures the inequality will hold for $p > \frac13$. The rest is algebra and manipulation as you noticed. – Macavity Mar 5 '14 at 6:20
@Macavity I am truly thankful for your help. It means a lot to me. I am on spring break and I can not get to my professor. Once again, thank you and r9m. – Frumpy Mar 5 '14 at 6:58

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.

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small typo in the last line .. the power should be $\frac {1}{p}$ not 3 – r9m Mar 5 '14 at 6:16
@r9m Yes. I have changed that. Could you explain to me exactly how I can go from the last few lines to the conclusion of the proof? I feel like there must be something missing. I don't have a particular reason, but it's just a feeling. – Frumpy Mar 5 '14 at 6:19
now its showing $p$, make it $\frac {1}{p}$, add the line $\large(\frac{a^p+b^p}{2})^{\frac{1}{p}} \ge \large(\frac{a^\frac{1}{3}+b^\frac{1}{3}}{2})^3$, for $p\ge \frac{1}{3}$ – r9m Mar 5 '14 at 6:25
@r9m Crap I see that. I'll fix it. You mean add that line prior to the generalized form correct? – Frumpy Mar 5 '14 at 6:29
@Frumpy Can you paste the LaTeX code at paste.ubuntu.com ? Then I will put it into this answer for you and maybe that will help you know how MathJax works. – 6005 Mar 6 '14 at 2:30

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 $E(2/3,1/3)\ge E(1,0)$

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., $\displaystyle (b-a) \le \left(\dfrac{a^{1/3}+b^{1/3}}{2}\right)^3(\ln b - \ln a)$.

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Yeah I'm only a junior math wise. I'm sure if I gave it some effort I could understand your proof, but my professor wanted us to go the way Mac did, so I will do so. Thank you very much for your help though, I always appreciate the help. – Frumpy Mar 5 '14 at 5:17
@user133156 It's Ok. I was trying to answer your first question .. " Why is it enough to prove this for the p=1/3 case only?" .. These are Generalized Stolasky Means ... I just gave an approach to proving the inequalities involving these means :) – r9m Mar 5 '14 at 5:27
Yeah I see that now, but I don't think our professor ever went over that haha. Thank you though :) I'm currently writing up the LaTex for the proof, if you'd like to look at it when I'm done. – Frumpy Mar 5 '14 at 5:43