# R and r When G on Incircle

### Solution

$IG=r.\,$ We know that (Geometric Inequalities, O. Bottema et al, p. 51)

$36(IG)^2=(a+b+c)^2+20r^2-64Rr.$

Thus, in our case, $4s^2=16r^2+64Rr,\,$ $s^2=4r^2+16Rr,\,$ where $2s=a+b+c.\,$ We assume as known that among all triangles with the same $R\,$ and $r\,$ the isosceles one has the shortest semiperimeter. If the latter is denoted $s_1,\,$ $s\ge s_1.$

If so, suffice it to prove that $s_1^2\ge 4r^2+16Rr.\,$ The Ravi substitution $a=y+z,\,$ $b=z+x,\,$ $c=x+y\,$ reduces the inequality to

$\displaystyle (x+y+z)^2\ge 4\frac{xyz}{x+y+z}+4\frac{(x+y)(y+z)(z+x)}{x+y+z},$

or, $(x+y+z)^3\ge 4xyz+4(x+y)(y+z)(z+x).\,$ The homogenization $x+y+z=1\,$ reduces this to

\displaystyle\begin{align} 1&\ge 4xyz+4(1-x)(1-y)(1-z)\\ &=4xyz+4(xy+yx+yz-xyz)=4(xy+yz+zx) \end{align}

and, say, $x=y=t,\,$ $z=1-2t\,$ for the isosceles triangle. Thus, we need to prove that $1\ge 4t^2+4(1-2t)\cdot 2t,\,$ or $12t^2-8t+1\ge 0.$

The polynomial $12t^2-8t+1\,$ has two roots $\displaystyle t_{1,2}=\frac{4\pm\sqrt{16-12}}{12}\,$ such that $t_1=\displaystyle \frac{1}{2},\,$ $t_2=\displaystyle \frac{1}{6}.\,$ The polynomial is not negative for $t\ge t_1\,$ and $t\le t_2.\,$ The former case is not possible as it leads to $z\le 0.\,$ Thus $t\le \displaystyle \frac{1}{6}.$

If $S\,$ denotes the area of $\Delta ABC,\,$

\displaystyle \begin{align} \frac{R}{r} &= \frac{abc}{4S}=\frac{abc\cdot s}{4s(s-a)(s-b)(s-c)}\\ &=\frac{(x+y)(y+z)(z+x)}{4xyz}\\ &=\frac{2t(1-t)^2}{4t^2(1-2t)}=\frac{(1-t)^2}{2t(1-2t)}=\frac{t^2-2t+1}{2t-4t^2}. \end{align}

Let $\displaystyle f(t)=\frac{t^2-2t+1}{2t-4t^2}.\,$ Then

\displaystyle \begin{align} f'(t) &= \frac{(2t-2)(2t-4t^2)-(2-8t)(t^2-2t+1)}{(2t-4t^2)^2}\\ &=\frac{(2t-2)(1-3t)}{(2t-4t^2)^2}\le 0, \end{align}

for $\displaystyle t\le\frac{1}{6}.\,$ It follows that

$\displaystyle f(t)\ge f\left(\frac{1}{6}\right)=\frac{25}{36}\cdot\frac{36}{12-4}=\frac{25}{8}.$

It follows that $\displaystyle \frac{R}{r}=f(t)\ge\frac{25}{8},\,$ i.e., $8R\ge 25r.$

### Acknowledgment

The problem by Leo Giugiuc and Kadir Altintas has been posted at the mathematical inequalities facebook page and solved by Marian Dinca who subsequently kindly communicated to me his solution.