An Inequality with Just Two Variables IV

If $a=2\,$ then the inequality becomes $\displaystyle 2+\frac{b+2}{b^2+b+1}\ge 2+1\,$ which is equivalent to $1\ge b^2,\,$ which is true, due to $b\in [0,1].\,$ Similarly, for $b=0.\,$ So assume $a,b\in (0,1].$

Consider the function $f:\,(-\infty,0]\to\mathbb{R},\,$ defined by $\displaystyle f(x)=\frac{e^x+2}{a^{2x}+e^x+1}.\,$ We have $\displaystyle f'(x)=-\frac{e^x(e^{2x}+e^x+1)}{(e^{2x}+e^x+1)^2}\,$ and $\displaystyle f''(x)=\frac{e^x(e^x-1)(e^{3x}+8e^{2x}+8e^x+1)}{(e^{2x}+e^x+1)^3}\le 0\,$ for $x\in (-\infty,0].\,$ It follows that $f\,$ is concave on $(-\infty,0].\,$ Denote $x=\ln a\,$ and $y=\ln b.\,$ Then $x+y\in [x+y,0]\,$ and $x+y=0+(x+y),\,$ making Karamata's inequality applicable: $f(x)+f(y)\ge f(0)+f(x+y),\,$ i.e.,

$\displaystyle\frac{a+2}{a^2+a+1}+\frac{b+2}{b^2+b+1}\ge \frac{ab+2}{(ab)^2+ab+1}+1.$

$\displaystyle\begin{align} &\frac{a+2}{a^2+a+1}+\frac{b+2}{b^2+b+1}- \frac{ab+2}{(ab)^2+ab+1}-1\\ &\qquad\qquad=\frac{1-a^2}{a^2+a+1}+\frac{1-b^2}{b^2+b+1}- \frac{1-(ab)^2}{(ab)^2+ab+1}-1\\ &\qquad\qquad=\frac{(1-a)(1-b)(1-ab)(1+a+ab)(1+b+ab)}{(a^2+a+1)(b^2+b+1)[(ab)^2+ab+1]}\\ &\qquad\qquad\ge 0, \end{align}$

for $a,b\in [0,1].$

The proof is by induction, with the original statement serving as the induction base, while the inductive step proceeds, with setting $\displaystyle a\prod_{k=1}^na_k\,$ and $b=a_{n+1}.$ We have

$\displaystyle\sum_{k=1}^n\frac{a_k+2}{a_k^2+a_k+1}\ge\frac{\displaystyle\prod_{k=1}^na_k+2}{\displaystyle\prod_{k=1}^na_k^2+\displaystyle\prod_{k=1}^na_k+1}+n-1$

and

$\displaystyle\frac{\displaystyle\prod_{k=1}^na_k+2}{\displaystyle\prod_{k=1}^na_k^2+\displaystyle\prod_{k=1}^na_k+1}+\frac{a_{n+1}+2}{a_{n+1}^2+a_{n+1}+1}\ge \frac{\displaystyle\prod_{k=1}^{n+1}a_k+2}{\displaystyle\prod_{k=1}^{n+1}a_k^2+\displaystyle\prod_{k=1}^{n+1}a_k+1}+1.$

Adding the two completes the induction.

The problem above has been kindly posted to the CutTheKnotMath facebook page by Leo Giugiuc; the result (and Solution 1) is due to Leo Giugiuc, Droberta Turnu Severin, Romania and (together with Solution 2) to Michael Rosenberg (Tel-Aviv, Israel). Originally, this was the basis for their inductive proof of a more general statement (Expansion).