, 2011

Let $D$, $E$, $F$ be the points where the incircle touches the sides $BC$, $CA$, $AB$, respectively, and let $x = AE = AF$, $y = BF = BD$, $z = CD = CE$.
Express all the lengths involved in the required inequality in terms of $x$, $y$ and $z$. Clearly, $AB = x+y$, $BC = y+z$, and $CA = z+x$. To express $AA_1$ and $AA_2$, use the similarity of the triangles $AA_1A_2$ and $ABC$. Their perimeters are $2x$ and $2(x+y+z)$, respectively, so $AA_1/AB = AA_2/AC = x/(x+y+z)$, whence $AA_1 = x(x+y)/(x+y+z)$ and $AA_2 = x(x+z)/(x+y+z)$. Similarly, $BB_1 = y(y+z)/(x+y+z)$, $BB_2 = y(y+x)/(x+y+z)$, $CC_1 = z(z+x)/(x+y+z)$ and $CC_2 = z(z+y)/(x+y+z)$.
We must show that
$$9 \sum x^2(x+y)(x+z) \geq (x+y+z)^2 \sum (x+y)^2.$$
Alternatively, but equivalently,
$$9 \sum x^4 + 3 \left(\sum x^2\right) \left(\sum xy\right) \geq 2 \left(\sum x^2\right)^2 + 4 \left(\sum xy\right)^2,$$
which is a consequence of the Cauchy-Schwarz inequality:
$$3(x^4 + y^4 + z^4) \geq (x^2 + y^2 + z^2)^2$$
and
$$x^2 + y^2 + z^2 \geq xy + yz + zx.$$
Clearly, equality holds if and only if $x = y = z$; that is, if and only if the triangle $ABC$ is equilateral.