Mathematical and Physical Journal
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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.

Statistics:

 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