An Inequality with Just Two Variables V

Problem

An Inequality  with  Just Two Variables V

Solution 1

Define function $\displaystyle f:\,\left(\frac{1}{x},1\right)\,\to\,\mathbb{R}\,$ by $\displaystyle \ln\left(\ln\frac{1}{x}\right).\,$ Then

$\displaystyle \begin{align} f'(x)&=\frac{1}{x\ln x}\\ f''(x)&=-\frac{\ln x+1}{(x\ln x)^2}\lt 0,\,\forall x\in\left(\frac{1}{e},1\right). \end{align}$

Hence, $f\,$ is concave on $\displaystyle \left(\frac{1}{e},1\right).\,$ By Jensen's inequality,

$\displaystyle f\left(\frac{b}{a+b}\cdot a+\frac{a}{a+b}\cdot b\right)\ge\frac{b}{a+b}f(a)+\frac{a}{a+b}f(a),$

i.e., $\displaystyle f\left(\frac{2ab}{a+b}\right)\ge\frac{bf(a)}{a+b}+\frac{af(b)}{a+b}.$ To continue,

$\displaystyle \ln\left(\ln\frac{a+b}{2ab}\right)\ge\frac{b}{a+b}\ln\left(\ln\frac{1}{a}\right)+\frac{a}{a+b}\ln\left(\ln\frac{1}{b}\right),$

i.e.,

$\displaystyle \ln\left(\ln\left(\frac{1}{2a}+\frac{1}{2b}\right)\right)\ge\ln\left(\left(\ln\frac{1}{a}\right)^{\frac{b}{a+b}}\cdot\left(\ln\frac{1}{b}\right)^{\frac{a}{a+b}}\right)$

which is

$\displaystyle \ln\left(\frac{1}{2a}+\frac{1}{2b}\right)\ge\left(\left(\ln\frac{1}{a}\right)^{b}\cdot\left(\ln\frac{1}{b}\right)^{a}\right)^{\frac{1}{a+b}}$

and, finally,

$\displaystyle \left(\ln\left(\frac{1}{2a}+\frac{1}{2b}\right)\right)^{a+b}\ge\left(\ln\frac{1}{a}\right)^{b}\cdot\left(\ln\frac{1}{b}\right)^{a}.$

Solution 2

Let $\displaystyle x=\frac{1}{a}\,$ and $\displaystyle y=\frac{1}{b};\,$ $x,y\in (1,e).\,$ We need to show that

$\displaystyle \ln\frac{x+y}{2}\ge\ln(x)^{\frac{x}{x+y}}\ln(y)^{\frac{y}{x+y}}.$

Let $y=x+\epsilon,\,$ where $0\lt\epsilon\lt e-x.\,$ We need to show that

$\displaystyle \ln\frac{2x+\epsilon}{2}\ge\ln(x)^{\frac{x}{2x+\epsilon}}\ln(x+\epsilon)^{\frac{x+\epsilon}{2x+\epsilon}}.$

Expanding up to orders of $\epsilon^2,\,$ $LHS=\ln +\displaystyle \frac{\epsilon}{2x}-\frac{\epsilon^2}{8x^2};\,$ $RHS=\ln x+\frac{\epsilon}{2x}-\frac{\epsilon^2}{8x^2\ln x}.\,$ So, since $\ln a\lt 1,\,$ we have $LHS \ge RHS.\,$

We can refine further with the higher orders of $O(\epsilon^4),\,$ by taking even orders. This is a more engineering oriented but functional approach.

Solution 3

Consider function $\displaystyle F(x)=x\ln (\ln (x)).\,$

$\displaystyle \begin{align} &F'(x)=\ln (\ln (x))+\frac{1}{\ln (x)},\\ &F''(x)=\frac{1}{x\ln (x)}\left[1-\frac{1}{\ln (x)}\right]. \end{align}$

As, for $x\in (1,e),\,$ $0\lt\ln (x)\lt 1,\,$ $F(x)\,$ is a concave function $(F''(x)\lt 0)\,$ for $x\,$ in this range. If so, for any two points $x,y\in (1,e),\,$

$\displaystyle f\left(\frac{x+y}{2}\right)\ge\frac{f(x)+f(y)}{2}.$

Applying this two $F(x)\,$ gives

$\displaystyle \frac{x+y}{2}\cdot\ln\left[\ln\left(\frac{x+y}{2}\right)\right]\ge\frac{x\ln [\ln (x)]+y\ln [\ln (x)]}{2},$

which, after simplification and exponentiation on both sides, yields the required inequality.

Acknowledgment

The problem above has been kindly posted to the CutTheKnotMath facebook page and several other forums (in particular at the Romanian Mathematics Magazine) by Dan Sitaru. After a length of time that it had not gathered any solution, Dan has communicated his solution by private mail. Solution 2 is by N. N. Taleb; Solution 3 is by Michel Charbonneau.

 

Inequalities in Two Variables

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