# 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