An Inequality with Inradius and Side Lengths

Problem

An Inequality with Inradius and Side Lengths - problem

Solution 1

$\displaystyle \begin{align}\sum_{cycl} (b+c-a)^2 &=4\sum_{cycl} a^2-\Bigr(\sum_{cycl} a\Bigr)^2\\ &=\sum_{cycl} a^2+\Bigr(3\sum_{cycl} a^2-\Bigr(\sum_{cycl} a\Bigr)^2\Bigr)\\ &\geq \sum_{cycl} a^2\overbrace{\geq}^{AM-GM} (3\sqrt[3]{abc})^2=9\sqrt[3]{(abc)^2}&(1)\\ \sum_{cycl} (b+c-a)^3&=\Bigr(\sum_{cycl} a\Bigr)^3-24abc\\ &\geq (3\sqrt[3]{abc})^3-24abc=27abc-24abc=3abc&(2) \end{align}$

From (1) and (2)

$\displaystyle\begin{align} &\sum_{cycl} (b+c-a)^2\cdot \sum_{cycl} (b+c-a)^3\geq 27\sqrt[3]{(abc)^5}\\ &\qquad\qquad=27\cdot (4RS)^{\frac{5}{3}}\geq 27\cdot \Bigr(4^5\cdot R^5\cdot (rs)^5\Bigr)^{\frac{1}{3}}\\ &\qquad\qquad\geq 27\Bigr(4^5\cdot (2r)^5\cdot (r\cdot 3\sqrt{3}r)^5\Bigr)^{\frac{1}{3}}\\ &\qquad\qquad=27\Bigr(8^5\cdot r^{15}\cdot (3\sqrt{3})^5\Bigr)^{\frac{1}{3}}\\ &\qquad\qquad=27\cdot 2^5\cdot (\sqrt{3})^5\cdot r^5=2592\sqrt{3}r^5. \end{align}$

Solution 2 and Illustrations

An intuitive (lazy) approach: let $a+b+c=1.\,$ If the LHS were convex with respect to the three variables (and it is, see illustrations), thw minimum would be reached for $\displaystyle s=a=b=c=\frac{1}{3},\,$ where the LHS equals $\displaystyle \frac{1}{27}.\,$ On the other hand, we know that an equilateral triangle has an inradius $\displaystyle r=\frac{s\sqrt{3}}{6}.\,$ So that the RHS equals $\displaystyle 2592(\sqrt{3})\left(\frac{1}{3}\cdot\frac{\sqrt{3}}{6}\right)^6=\frac{1}{27}.$

Taleb's illustration 1

Or this, easier to see.

Taleb's illustration 2

Acknowledgment

I am grateful to Dan Sitaru for kindly posting a problem of his from the Romanian Mathematical Magazine at the CutTheKnotMath facebook page and later mailing me his solution on a LaTeX file. Solution 2 with illustrations is by N. N. Taleb.

 

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