21st Mediterranean Mathematical Competition

It is well-known that in an equilateral triangle the sum of the distances from an interior point $P$ to its sides equals the altitude of the triangle, as can be easily proven. On account of the preceding, we have to prove that if $x+y+z=1$ then it holds that $$x^2+y^2+z^2\ge x^3+y^3+z^3+6xyz$$ To do it, we begin observing that when $x+y+z=1$, then $$xy+yz+zx\ge 9xyz$$ Indeed, applying AM-GM inequality, we get $$xy+yz+zx=(xy+yz+zx)(x+y+z)\ge 3\sqrt[3]{(xy)(yz)(zx)}\cdot 3\sqrt[3]{xyz}=9xyz$$ or $$xy+yz+zx-3xyz\ge 6xyz$$ On account of the constrain and the preceding inequality, we obtain $$\begin{array}{rl}xy+yz+zx-3xyz& =xy(1-z)+yz(1-x)+zx(1-y)\\ & =xy(x+y)+yz(y+z)+zx(z+x)\\ & \ge 6xyz\end{array}$$ Adding $1$ to both terms of the last inequality and reordering terms, yields $$(x+y+z{)}^2+xy(x+y)+yz(y+z)+zx(z+x)-6xyz\ge 1$$ or equivalently, $$x^2+y^2+z^2+2xy(1-z)+2yz(1-x)+2zx(1-y)+xy(x+y)+yz(y+z)+zx(z+x)\ge 1,$$ and $$x^2+y^2+z^2+3xy(x+y)+3yz(y+z)+3zx(z+x)\ge 1=(x+y+z{)}^3$$ from which $$x^2+y^2+z^2\ge x^3+y^3+z^3+6xyz$$ follows. Equality holds when $x=y=z=1/3$. That is, when $P$ is the centroid of the triangle, and we are done.