Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
Already signed up?
New to KöMaL?

Problem A. 623. (October 2014)

A. 623. Let \(\displaystyle a\), \(\displaystyle b\) and \(\displaystyle c\) be three distinct positive reals. The logarithmic mean of \(\displaystyle a\), \(\displaystyle b\), \(\displaystyle c\) is defined by

\(\displaystyle L(a,b,c) = 2\left( \frac{a}{(\ln a-\ln b)(\ln a-\ln c)} + \frac{b}{(\ln b-\ln c)(\ln b-\ln a)} + \frac{c}{(\ln c-\ln a)(\ln c-\ln b)} \right).\)

Prove that \(\displaystyle \sqrt[3]{abc} < L(a,b,c) < \frac{a+b+c}{3}\).

(5 pont)

Deadline expired on November 10, 2014.

Solution. First we prove the known formula

\(\displaystyle L(a,b,c) = 2 \mathop{\int\int}\limits_{\scriptsize\begin{matrix} x,y\ge0,\\ x+y\le1 \end{matrix}} a^x b^y c^{1-x-y} \,{\rm d}x{\rm d}y. \)(1)

Due to the homogenity, \(\displaystyle L(a,b,c)=c\cdot L(a/c,b/c,1)\), it suffices to verify (1) in the particular case \(\displaystyle c=1\):

\(\displaystyle \mathop{\int\int}\limits_{x+y\le 1} a^x b^y \,{\rm d}x{\rm d}y % = \int_{x=0}^1 a^x \left( \int_{y=0}^{1-x} b^y \,{\rm d}y \right) \,{\rm d}x % = \int_0^1 a^x \frac{b^{1-x}-1}{\log b} \,{\rm d}x = \frac1{\log b} \int_0^1 \left( b\left(\frac{a}{b}\right)^x - a^x \right) \,{\rm d}x \)

\(\displaystyle = \frac1{\log b} \left( \frac{a-b}{\log\frac{a}{b}} -\frac{a-1}{\log a}\right) % = \frac{a}{\log b} \left(\frac1{\log\frac{a}{b}}-\frac1{\log a}\right) -\frac{b}{\log b\cdot\log\frac{a}{b}} +\frac1{\log a\cdot \log b} % \\= \frac{a}{(\log a -\log b)\log a} + \frac{b}{(\log b -\log a)\log b} + \frac1{\log a \cdot \log b} = \frac12 L(a,b,1). \)

The identity (1) is proved.

Now apply the AM-GM inequality to the integrand in (1):

\(\displaystyle L(a,b,c) % = 2 \mathop{\int\int}\limits_{x+y\le1} a^x b^y c^{1-x-y}\,{\rm d}x{\rm d}y % \le 2 \mathop{\int\int}\limits_{x+y\le1} \big(xa+yb+(1-x-y)c\big)\,{\rm d}x{\rm d}y = \frac{a+b+c}{3}. \)

Then, applying Jensen's inequality to the exponential function, we get

\(\displaystyle L(a,b,c) % = 2\mathop{\int\int}\limits_{x+y\le1} \exp\big( x\log a + y\log b + (1-x-y)\log c \big) \,{\rm d}x{\rm d}y \ge % \)

\(\displaystyle \ge \exp\left(2\mathop{\int\int}\limits_{x+y\le1} \big( x\log a + y\log b + (1-x-y)\log c \big) \,{\rm d}x{\rm d}y \right) =% \)

\(\displaystyle = {\rm exp} \frac{\log a+\log b+\log c}{3} = \root3\of{abc}. \)

Remark. For \(\displaystyle n\) positive numbers \(\displaystyle a_1,a_2,\ldots,a_n\), the logarithmic mean can be defined as

\(\displaystyle L(a_1,a_2,\ldots,a_n) = (n-1)! \cdot \exp[\log a_1,\log a_2,\ldots, \log a_n] \)

where \(\displaystyle \exp[\ldots]\) denotes the divided difference of the exponential function. It is known that

\(\displaystyle L(a_1,a_2,\ldots,a_n) = (n-1)! \cdot \mathop{\int\ldots\int}\limits_{\scriptsize\begin{matrix} x_1,\ldots,x_{n-1}\ge0,\\ x_1+\ldots+x_{n-1}\le1 \end{matrix}} a_1^{x_1} a_2^{x_2} \cdots a_{n-1}^{x_{n-1}} a_n^{1-x_1-\ldots-x_{n-1}} \,\mathrm{d}x_1\cdots\mathrm{d}x_{n-1}. \)

From this formula, the relation \(\displaystyle GM\le LM\le AM\) can be derived in the same way as in the solution.


8 students sent a solution.
5 points:Fehér Zsombor, Janzer Barnabás, Saranesh Prembabu, Szabó 789 Barnabás, Williams Kada.
4 points:Di Giovanni Márk.
2 points:2 students.

Problems in Mathematics of KöMaL, October 2014