Geometric (Trigonometric) inequality $\frac{(a+b+c)^3}{3abc}\leq1+\frac{4R}{r}$

Question

How can one prove/disprove that $\frac{(a+b+c)^3}{3abc}\leq1+\frac{4R}{r}$ where $R$ and $r$ denote the usual circum and inradii respectively.

I know that $R=\frac{abc}{4\Delta}$ and $r=\frac{\Delta}{s}$, where $\Delta$ denotes area of triangle, and $s$ the semi perimeter. Any ideas. Thanks beforehand.

This is problem J392 from the Problem column of Mathematical Reflections - Issue 6 2016.

Answer 1

We can prove that $$\frac{(a+b+c)^3}{3abc}\leq1+\frac{4R}{r}$$ Indeed, we need to prove that $$\frac{(a+b+c)^3}{3abc}\leq1+\frac{\frac{abc}{S}}{\frac{2S}{a+b+c}}$$ or $$\frac{(a+b+c)^3}{3abc}\leq1+\frac{8abc(a+b+c)}{16S^2}$$ or $$\frac{(a+b+c)^3}{3abc}\leq1+\frac{8abc}{\prod\limits_{cyc}(a+b-c)}$$ or $$\frac{(a+b+c)^3}{3abc}\leq\frac{\sum\limits_{cyc}(-a^3+a^2b+a^2c+2abc)}{\prod\limits_{cyc}(a+b-c)}$$ or $$\frac{(a+b+c)^3}{3abc}\leq\frac{\sum\limits_{cyc}(-a^3+abc+a^2b+a^2c+abc)}{\prod\limits_{cyc}(a+b-c)}$$ $$\frac{(a+b+c)^2}{3abc}\leq\frac{\sum\limits_{cyc}(-a^2+ab+ab)}{\prod\limits_{cyc}(a+b-c)}$$ or $$\sum\limits_{cyc}a\sum\limits_{cyc}(2a^2b^2-a^4)\leq3\sum\limits_{cyc}(2a^2b^2c-a^3bc)$$ or $$2\sum\limits_{cyc}(a^4b+a^4c-a^3b^2-a^3c^2-2a^3bc+2a^2b^2c)+\sum\limits_{cyc}(a^5-a^4b-a^4c+a^3bc)\geq0,$$ for which it's enough to prove that $$\sum\limits_{cyc}(a-b)^2(ab(a+b)-abc)\geq0$$ or $$\sum\limits_{cyc}(a-b)^2ab(a+b-c)\geq0.$$ Done!