A Problem from the Danubius Contest 2016

Problem

Dorin Marghidanu has kindly posted at the CutTheKnotMath facebook page the following problem, along with a solution (Solution 1.) The problem appeared at the 2016 Danubius Contest.

A Problem from the Danubius Contest 2016

Solution 1

Introducing $\displaystyle x=\frac{a}{b},\;$ we have $x\gt 0\;$ and $\displaystyle E(a,b)=\frac{2x+3}{4x+5}+\frac{2+3x}{4+5x}.\;$

For the function $f:\;(0,\infty)\rightarrow\mathbb{R},\;$ where $\displaystyle f(x)=\frac{2x+3}{4x+5}+\frac{2+3x}{4+5x},\;$ we have $\displaystyle f'(x)=\frac{-18(x^2-1)}{(4x+5)^2(4+5x)^2},\;$ so that $f'(x)=0\;$ only for $x=1.\;$ Also $f'(x)\gt 0\;$ for $x\lt 1\;$ and $f'(x)\lt 0\;$ for $x\gt 1,\;$ making $x=1\;$ the only point of local maximum. $\displaystyle f(1)=\frac{10}{9},\;$ implying $\displaystyle f(x)\le\frac{10}{9}.$

Function $f\;$ is increasing on $(0,1)\;$ and decreasing on $(1,\infty),\;$ with

$\displaystyle\lim_{x\rightarrow 0^+}=\frac{3}{5}+\frac{2}{4}=\frac{11}{10}=\frac{1}{2}+\frac{3}{5}=\lim_{x\rightarrow \infty}.$

Which means that $\displaystyle f(x)\gt\frac{11}{10}.\;$ Thus $\displaystyle f:\;(0,\infty)\rightarrow \left(\frac{11}{10},\frac{10}{9}\right].\;$ We may say more: since the function $f\;$ is continuous, it takes on all the intermediate values, i.e., $\displaystyle f\big((0,\infty)\big)=\left(\frac{11}{10},\frac{10}{9}\right].$

Solution 2

As above, introduce $\displaystyle x=\frac{a}{b}\;$ and note that $\displaystyle E(a,b)=1+\frac{2x^2+5x+2}{20x^2+41x+20}=1+\frac{2(x+1)^2+x}{20(x+1)^2+x}.\;$ Verify now directly that $\displaystyle\frac{1}{9}\ge\frac{2(x+1)^2+x}{20(x+1)^2+x}\gt\frac{1}{10}.\;$ Indeed, the left inequality is equivalent to $(x-1)^2\ge 0,\;$ with equality for $x=1,\;$ while the right inequality is equivalent to $10x\gt x,\;$ which becomes an equality as $x\rightarrow 0^+.$

Illustration

TalebDanubius2016

Acknowledgment

Dorin Marghidanu has kindly posted at the CutTheKnotMath facebook page the following problem, along with a solution (Solution 1.) The problem appeared at the 2016 Danubius Contest. The illustration is by Nassim Nicholas Taleb.

 

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