An Inequality in Triangle and in General

Remark

Both solutions below use the fact that in any triangle

$\cot A\cot B+\cot B\cot C+\cot C\cot A = 1.$

Thus, using the substitution $a=\cot A,\,$ $b=\cot B,\,$ $c=\cot C\,$ the problem reduces to proving that

Prove that for positive $a,b,c\,$ such that $ab+bc+ca=1,$

$\displaystyle\sum_{cycl}\frac{ab^3}{b^2+2a^2}+2\sum_{cycl}\frac{a^2b}{a+2b}\ge 1.$

Solution 1

The required inequality is equivalent to

$\displaystyle\sum_{cycl}ab-2\sum_{cycl}\frac{a^3b}{b^2+2a^2}+2\sum_{cycl}\frac{a^2b}{a+2b}\ge 1,$

(1)

$\displaystyle\sum_{cycl}\frac{a^2b}{a+2b}\ge \sum_{cycl}\frac{a^3b}{b^2+2a^2}.$

We'll prove $\displaystyle\frac{a^2b}{a+2b}\ge \frac{a^3b}{b^2+2a^2}\,$ which is equivalent to $2a^2+b^2\ge a(a+2b),\,$ i.e., $(a-b)^2\ge 0,\,$ implying (1).

Equality holds when $a=b=c=\displaystyle\frac{1}{\sqrt{3}},\,$ i.e., when $A=B=C=60^{\circ}.$

Solution 2

\displaystyle\begin{align} \frac{ab^3}{b^2+2a^2}+2\frac{a^2b}{a+2b} &= ab\left(\frac{b^2}{b^2+2a^2}+\frac{(2a)(2a)}{(a+2b)(2a)}\right)\\ &=ab\left(\frac{b^2}{b^2+2a^2}+\frac{(2a)^2}{2a^2+4ab}\right)\\ &\ge ab\left(\frac{(b+2a)^2}{(b+2a)^2}\right)\\ &= ab, \end{align}

where, on the penultimate step, we used Bergströms' inequality. Summing up and using $ab+bc+ca=1,\,$ delivers the required inequality.

Acknowledgment

Dan Sitaru has kindly posted the above problem (from his book "Math Accent") at the CutTheKnotMath facebook page. Solution 1 is by Dung Thanh Tùng; Solution 2 is by Myagmarsuren Yadamsuren.