Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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 at 2:00
    
Any hint would be helpful. I am stuck and I can not get past this. –  Frumpy Mar 5 at 2:37

3 Answers 3

up vote 7 down vote accepted

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

share|improve this answer
    
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 at 4:12
    
@Macavity +1 Nice :) –  r9m Mar 5 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 at 4:27
1  
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 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 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.

enter image description here

enter image description here

enter image description here

share|improve this answer
1  
small typo in the last line .. the power should be $\frac {1}{p}$ not 3 –  r9m Mar 5 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 at 6:19
1  
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 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 at 6:29
1  
@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. –  Goos Mar 6 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 $E(2/3,1/3)\ge E(1,0)$

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

share|improve this answer
    
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 at 5:17
1  
@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 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 at 5:43

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.