parmenides51 wrote: Let $P$ be a point in the interior of an equilateral triangle with height $1$, and let $x,y,z$ denote the distances from $P$ to the three sides of the triangle. Prove that \[ x^2+y^2+z^2 ~\ge~ x^3+y^3+z^3 +6xyz \] We have $x+y+z=1$. Therefore: $$x^2+y^2+z^2=(x+y+z)(x^2+y^2+z^2) = x^3+y^3+z^3+\sum_{cyc}x^2(y+z) \geq x^3+y^3+z^3+6xyz$$