Đặt $x=\frac{1}{a}, \ \ y=\frac{1}{b}, \ \ z=\frac{1}{c}$,
Bất đẳng thức $\Leftrightarrow \sum_{cyc}\frac{9x^2yz}{x^2+2yz}\leq (x+y+z)^2$,
$$\sum_{cyc}(x^2+2yz-\frac{9x^2yz}{x^2+2yz})\geq 0$$
$$\sum_{cyc}\frac{(x^2-4yz)((x+y)(x-z)+(x+z)(x-y))}{x^2+2yz}\geq 0$$
$$\sum_{cyc}(z^2+2xy)[(x^2-4yz)(x+z)(y^2+2zx)-(y^2-4zx)(y+z)(x^2+2yz)](x-y)\geq 0$$
$$\sum_{cyc}S_z(x-y)^2\geq 0$$
Mà $S_x,S_y,S_z\ge 0$. đpcm.