Let $r_a$ be the radius of the largest escribed circle of a triangle, and let $R$ denote the radius of the circumscribed circle. Prove that $2r_a \geq 3R$.
How to prove it? I am not keen on inequalities, but I think (since it is from Paul Erdős) it is a beautiful problem. Please help! Thank you!
Let the circumradius of the triangle be $R$, then $a=2R\sin(A)$ (this can be proven using the Inscribed Angle Theorem) and we also have that the area of the triangle is $\Delta=\frac12bc\sin(A)$. Thus, $$ abc=2R\sin(A)bc=4R\Delta\tag1 $$ Therefore, using the Law of Cosines to get $\cos(A)=\frac{b^2+c^2-a^2}{2bc}$, we have $$ \begin{align} \tan(A/2) &=\frac{\sin(A)}{1+\cos(A)}\\[6pt] &=\frac{\frac{a}{2R}}{1+\frac{b^2+c^2-a^2}{2bc}}\\ &=\frac{\frac{abc}{R}}{(b+c)^2-a^2}\\[9pt] &=\frac\Delta{s(s-a)}\tag2 \end{align} $$ where $s=\frac{a+b+c}2$.
Assume that $a\ge b\ge c$ and let $r_a$ be the exradius on the side of length $a$.
In the diagram below, $\angle CBD=\pi-B$; thus, $\angle CBO=\frac\pi2-\frac B2$, and therefore, $\angle FOB=\frac B2$. Similarly, $\angle FOC=\frac C2$.
Note that $$ a=r_a\tan(B/2)+r_a\tan(C/2)\tag3 $$ and therefore, applying the analogs of $(2)$ for $B$ and $C$, $$ \begin{align} r_a &=\frac{a}{\tan(B/2)+\tan(C/2)}\\[6pt] &=\frac{a}{\frac\Delta{s(s-b)}+\frac\Delta{s(s-c)}}\\[6pt] &=\frac{s(s-b)(s-c)}\Delta\tag4 \end{align} $$ From $(1)$, we have $$ R=\frac{abc}{4\Delta}\tag5 $$ Dividing $(5)$ by $(4)$ yields $$ \begin{align} \frac{R}{r_a} &=\frac{abc}{4s(s-b)(s-c)}\\[6pt] &=\frac{bc}{4s(s-c)}+\frac{bc}{4s(s-b)}\tag6 \end{align} $$ For a particular $s$, each of the summands in $(6)$ increase in $b$ and $c$. Since $a\ge b\ge c$, we have the maximum when $a=b=c=\frac23s$, that is $$ \frac{R}{r_a}\le\frac23\tag7 $$