When Is Triangle Equilateral:
Miguel Ochoa Sanchez's Criterion

First observe that the identity in the problem is equivalent to $\displaystyle ax^2+by^2+cz^2=\frac{5}{4}abc.\;$ Now, using the barycentric coordinates, the circle with center $M(u,v,w),\;$ $u+v+w=1,\;$ and radius $r,\;$ has the equation

$ux^2+vy^2+wz^2 = r^2+a^2vw+b^2wu+c^2uv.$

With this in mind, the incircle is described by

$\displaystyle \begin{align}\frac{1}{a+b+c}\left(ax^2+by^2+cz^2\right)&=r^2+\frac{1}{(a+b+c)^2}(a^2bc+ab^2c+abc^2)\\ &=r^2+\frac{abc}{a+b+c}. \end{align}$

Thus the problem reduces to showing that

$\displaystyle r^2(a+b+c)+abc=\frac{5}{4}abc,$

which is the same as

$4r^2(a+b+c)=abc.$

However, $abc=4RS,\;$ where $R\;$ is the circumradius and $S\;$ the area of $\Delta ABC.\;$ If $s\;$ is the semiperimeter of the triangle then $rs=S\;$ and $4r^2(a+b+c)=8rS,\;$ meaning that the identity is equivalent to $2r=R,\;$ which only holds for the equilateral triangle.

The problem, due to Miguel Ochoa Sanchez, has been kindly posted by Leo Giugiuc at the CutTheKnotMath facebook page.