Problem 1567

Let $a = \frac{1}{2}x$, $b = y$, $c = \frac{1}{3}z$, then the left-hand side of the problem can be transformed into
$$\frac{2x^2 + xy}{(y + \sqrt{xz} + z)^2} + \frac{2y^2 + yz}{(z + \sqrt{xy} + x)^2} + \frac{2z^2 + xz}{(x + \sqrt{yz} + y)^2} \geq 1.$$
By the Cauchy-Schwarz inequality we can obtain
$$(yx + x^2 + z^2) \left( \frac{y}{x} + \frac{z}{x} + \frac{z^2}{x^2} \right) \geq (y + \sqrt{xz} + z)^2.$$
Therefore we obtain
$$\frac{2x^2 + xy}{(y + \sqrt{xz} + z)^2} \geq \frac{x^2}{xy + xz + z^2},$$
and equality in this expression holds if and only if $x = z$.
Replacing $(x, y, z)$ with $(y, z, x)$ and $(z, x, y)$ respectively, we obtain
$$\frac{2y^2 + yz}{(z + \sqrt{xy} + x)^2} \geq \frac{y^2}{yz + yx + x^2},$$
and
$$\frac{2z^2 + xz}{(x + \sqrt{yz} + y)^2} \geq \frac{z^2}{zx + zy + y^2}.$$
(Equality holds respectively if and only if $y = x$ and $z = y$.)
Let $M = \frac{x^2}{xy+xz+z^2} + \frac{y^2}{yz+yx+x^2} + \frac{z^2}{zx+zy+y^2}$, and using the Cauchy-Schwarz inequality again we obtain
$$M((xy + xz + z^2) + (yz + yx + x^2) + (zx + zy + y^2)) \geq (x + y + z)^2.$$
Since
$$(xy + xz + z^2) + (yz + yx + x^2) + (zx + zy + y^2) = (x + y + z)^2,$$
we immediately obtain
$$M = \frac{x^2}{xy + xz + z^2} + \frac{y^2}{yz + yx + x^2} + \frac{z^2}{zx + zy + y^2} \geq 1,$$
and equality in this expression holds if and only if $x = y = z$. That is,
$$\frac{8a^2 + 2ab}{(b + \sqrt{6ac} + 3c)^2} + \frac{2b^2 + 3bc}{(3c + \sqrt{2ab} + 2a)^2} + \frac{18c^2 + 6ac}{(2a + \sqrt{3bc} + b)^2} \geq 1,$$
and equality in this expression holds if and only if $2a = b = 3c$.