An Inequality with Powers And Logarithm

Problem

An  Inequality with Powers And Logarithm, problem

Solution

Consider function $f:\,[1,\infty)\to \mathbb{R};\,$ $\displaystyle f(x)=x-\frac{1}{x}-2\ln x.$

$\displaystyle f'(x)=1+\frac{1}{x^2}-\frac{2}{x}=\frac{x^2-2x+1}{x^2}=\frac{(x-1)^2}{x^2}\geq 0.$

It follows that the function is increasing: $f(x)\ge f(1)=0,\,$ for $x\ge 1.\,$ In other words, $\displaystyle x-\frac{1}{x}\ge 2\ln x.$

We'll use this result with $\displaystyle x=\frac{a}{b},\frac{a^2}{b^2},\frac{a^3}{b^3}\,$ to obtain

$\displaystyle \begin{align} \frac{a}{b}-\frac{b}{a}&\geq 2\ln \Bigr(\frac{a}{b}\Bigr)\\ \frac{a^2}{b^2}-\frac{b^2}{a^2}&\geq 2\ln \Bigr(\frac{a}{b}\Bigr)^2=4\ln \Bigr(\frac{a}{b}\Bigr)\\ \frac{a^3}{b^3}-\frac{b^3}{a^3}&\geq 2\ln \Bigr(\frac{a}{b}\Bigr)^3=6\ln \Bigr(\frac{a}{b}\Bigr). \end{align}$

Adding up and rearranging we get

$\displaystyle \frac{a}{b}+\frac{a^2}{b^2}+\frac{a^3}{b^3}\ge\frac{b}{a}+\frac{b^2}{a^2}+\frac{b^3}{a^3}+ 12 \ln \Bigr(\frac{a}{b}\Bigr),$

or,

$\displaystyle \frac{a}{b}+\frac{a^2}{b^2}+\frac{a^3}{b^3}+12\ln b\ge\frac{b}{a}+\frac{b^2}{a^2}+\frac{b^3}{a^3}+12\ln a.$

Extra

I originally misread the problem as

Prove, for $a,b\gt 0,\,$ the following inequality

$\displaystyle \frac{a}{b}+\frac{a^2}{b^2}+\frac{a^3}{b^3}+12\ln b\ge\frac{b}{a}+\frac{b^2}{a^2}+\frac{b^3}{a^3}+12\ln a.$

Find a simple argument to show that, as stated, it could not be true.

Acknowledgment

Dan Sitaru has kindly posted at CutTheKnotMath facebook page a problem of his from the Romanian Mathematical Magazine and later sent me a LaTeX file with his solution.

Concerning Extra

If the left-hand side is denoted $f(a,b),\,$ the right-hand side becomes $f(b,a)\,$ and the misread problem suggests that $f(a,b)\ge f(b,a),\,$ for $a,b\gt 0.\,$ The problem allows one to swap variables and claim $f(b,a)\ge f(a,b)\,$ which, in combination, leads to $f(a,b)=f(b,a)\,$ and this is patently not true.

 

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