An Inequality of Degree 3 with Inradius

Problem

An Inequality of Degree 3 with Inradius - problem

Solution 1

$\displaystyle \sum_{cycl}(a+b-c)^3=8\sum_{cycl}(p-a)^3$ where $2p=a+b+c.$

Let $x=p-a,$ $y=p-b,$ $z=p-c.$ Then $\displaystyle \sum_{cycl}(p-a)(p-b)=r(4R+r),$ where $R$ is the circumradius of $\Delta ABC,$ and $\displaystyle \prod_{cycl}(p-a)=pr^2.$ Further.

$\displaystyle \begin{align} \sum_{cycl}x^3 &= \left(\sum_{cycl}x\right)^3-3\left(\sum_{cycl}x\right)\left(\sum_{cycl}xy\right)+3xyz\\ &=p^3-3p\cdot r(4R+r)+3pr^2=p^3-12Rrp. \end{align}$

So,

$\displaystyle \begin{align} \displaystyle \sum_{cycl}(a+b-c)^3+24abc&=8\sum_{cycl}(p-a)^3+96Rrp\\ &=8(p^3-12Rrp)+96Rrp=8p^3\\ &\ge 8\cdot (3\sqrt{3}r)^3=648\sqrt{3}r^3. \end{align}$

Solution 2

By Chebyshev's inequality,

$\displaystyle \begin{align} \sum_{cycl}(a+b-c)^3 &\ge\frac{1}{9}\left(\sum_{cycl}(a+b-c)\right)^3\\ &=\frac{1}{9}\cdot (a+b+c)^3=\frac{8}{9}\cdot p^3. \end{align}$

Thus, using $p\ge 3\sqrt{3}r$ and $R\ge 2r,$

$\displaystyle\begin{align} \sum_{cycl}(a+b-c)^3+24abc &\ge \frac{8}{9}p^3+96pRr\\ &=8p\left(\frac{1}{9}\cdot 27r^2+24r^2\right)\\ &=27\cdot 24\cdot\sqrt{3}\cdot r^3\\ &=648\sqrt{3}r^3. \end{align}$

Acknowledgment

Dan Sitaru has kindly posted the above problem of his, with solutions, at the CutTheKnotMath facebook page. The problem was originally published at the Romanian Mathematical Magazine. Solution 1 is by Diego Alvariz (Soumava Chakraborty gave pratically the same solution); Solution 2 is by Yadamsuren Myagmarsuren.

 

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