Let $D, E$ and $F$ be the common points of with the sides $BC, CA$ and $AB$, respectively, and let $AE = AF = x$, $BF = BD = y$ and $CD = CE = z$.
Since $\triangle AA_1A_2 \sim \triangle ABC$, then $\frac{AA_1}{AB} = \frac{AA_2}{AC} = \frac{P_{AA_1A_2}}{P_{ABC}} = \frac{x}{x+y+z}$ and hence $AA_1 = \frac{x(x+y)}{x+y+z}$ and $AA_2 = \frac{x(x+z)}{x+y+z}$. Analogously, $BB_1 = \frac{y(y+z)}{x+y+z}$, $BB_2 = \frac{y(y+x)}{x+y+z}$, $CC_1 = \frac{z(z+x)}{x+y+z}$ and $CC_2 = \frac{z(z+y)}{x+y+z}$.
Then the given inequality is equivalent to
$$9 \sum x^2 (x+y)(x+z) \ge (x+y+z)^2 \sum (x+y)^2,$$
i.e.
$$9 \sum x^4 + 2(\sum x^2)(\sum xy) \ge 2(\sum x^2)^2 + 4(\sum xy)^2,$$

The last inequality follows by the well-known inequalities
$$3(x^4 + y^4 + z^4) \geq (x^2 + y^2 + z^2)^2 \text{ and } x^2 + y^2 + z^2 \geq xy + yz + zx.$$