# Proving an inequality regarding the surfaces of triangles

Let $$S_{\mathcal T}$$ be the surface of triangle $$\mathcal T$$. Let $$3$$ isometrical circles meet at a certain point, and name $$A$$, $$B$$, $$C$$ their $$3$$ pairwise intersections. Let $$XYZ$$ be the triangle containing the $$3$$ circles such that each side is tangent to $$2$$ of them. Show that $$S_{XYZ}\ge 9S_{ABC}$$

After thinking about this problem for a bit, and looking at a drawing of it, the circumradius of $$XYZ$$ seems to be $$r_1+r_2$$, where $$r_1$$ is the common radius of the $$3$$ isometrical circles and $$r_2$$ is the inradius of the triangle formed by the centers of the circles (which I believe is also the inradius of $$ABC$$, since both triangles seem isometrical). If this is proven, then we can easily conclude with Euler's relation...

Let us assume that the radius of each original circle is $$1$$, and let us denote with $$D,E,F$$ the centers of such circles and with $$P$$ their common point.
A homotethy with center $$P$$ and factor $$\frac{1}{2}$$ maps $$ABC$$ into the medial triangle of $$DEF$$, hence $$[ABC]=[DEF]$$. $$XYZ$$ and $$DEF$$ have parallel sides and the inradius of $$XYZ$$ is just one plus the inradius of $$DEF$$, so
$$\frac{[XYZ]}{[ABC]}=\frac{[XYZ]}{[DEF]}=\left(\frac{r_{DEF}+1}{r_{DEF}}\right)^2$$ and it is enough to show that $$r_{DEF}\leq \frac{1}{2}$$. On the other hand the circumradius of $$DEF$$ is $$1=PD=PE=PF$$ and by Euler's inequality $$r_{DEF}\leq \frac{1}{2}R_{DEF}$$: we are done.
Since $$DEF$$ and $$ABC$$ are isometrical, we have also proved Johnson's theorem $$R_{ABC}=1$$.