Seems easy huh?

This problem can be approached in numerous ways once RMS $\ge$ AM $\ge$ GM $\ge$ HM is understood. (RMS = root-mean square, AM = arithmetic mean, GM = geometric mean, HM = harmonic mean)

In other words: $\sqrt{\frac{x_1^2+\cdots+x_n^2}{n}} \ge\frac{x_1+\cdots+x_n}{n}\ge\sqrt[n]{x_1\cdots x_n}\ge\frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}}$

From here, we can see that: $\frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \le \frac{a+b+c}{3}$

It becomes very clear now that both $a+b+c$ and $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ must be 3. In addition, this is only satisfied when $a$ , $b$ , $c$ , are all equal, which confirms that they are each equal to 1. The rest of the problem is trivial.

To check, we apply the same relations to the following: $\sqrt{\frac{a^2+b^2+c^2}{3}} \ge \frac{a+b+c}{3} \\ \sqrt{\frac{a^2+b^2+c^2}{3}} \ge 1 \\ a^2+b^2+c^2 \ge 3$

This tells us that $(a+b+c)^2 = a^2 + b^2 +c^2 + 2(ab+bc+ac) = 9 \\ a^2 + b^2 + c^2 = 9-2(ab+bc+ac) \ge 3 \\ ab+bc+ac \ge 3$

Both of which hold true if all the variables are 1.

From here, we could expand out the formula and find more relations and simplify, but it is far easier to just plug in the value of 1 for each of the variables, confirming the answer is $\boxed{\frac{3}{16}}$ , and no inequality relation is even necessary.

Note: I personally think the problem should be worded a bit more technically, specifying that $\frac{x}{y}$ should be the minimum value possible.

For positive real numbers $x, y, z,$ from the arithmetic-geometric-mean inequality,

$2x+y+z=(x+y)+(x+z)\geq2\sqrt{(x+y)(y+z)},$ we obtain $\dfrac{1}{(2x+y+z)^2}\leq\dfrac{1}{4(x+y)(x+z)}.$

Applying this to the left-hand side terms of the inequality to find, we get

$\dfrac{1}{(2a+b+c)^2}+\dfrac{1}{(2b+a+c)^2}+\dfrac{1}{(2c+a+b)^2}\leq$ $\dfrac{1}{4(a+b)(a+c)}+\dfrac{1}{4(b+c)(b+a)}+\dfrac{1}{4(c+a)(c+b)}=$

$\dfrac{(b+c)+(c+a)+(a+b)}{4(a+b)(b+c)(a+c)}=\dfrac{a+b+c}{2(a+b)(b+c)(a+c)}.............\boxed{1}$

A second application of the inequality of the arithmetic-geometric mean yields,

$a^2b+ab^2+b^2c+bc^2+a^2c+ac^2\geq6abc$ or, equivalently,

$9(a+b)(b+c)(c+a)\geq8(a+b+c)(ab+bc+ac).................\boxed{2}$

The supposition $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=a+b+c,$ can be written as

$ab+bc+ac=abc(a+b+c)........................................\boxed{3}$

Applying the arithmetic-geometric-mean inequality

$x^2y^2+x^2z^2\geq2x^2yz$ thrice, we get

$a^2b^2+b^2c^2+c^2a^2\geq a^2bc+ab^2c+abc^2,$ which is equivalent to,

$(ab+bc+ca)^2\geq 3abc(a+b+c)...............................\boxed{4}$

Combining $\text{(1), (2), (3), and (4)}$ , we will finish the solution:

$\dfrac{a+b+c}{2(a+b)(b+c)(a+c)}=\dfrac{(a+b+c)(ab+bc+ac)}{2(a+b)(b+c)(a+c)}*\dfrac{ab+bc+ac}{abc(a+b+c)}*\dfrac{{abc(a+b+c)}}{(ab+bc+ac)^2}$

$\leq \dfrac{9}{2*8}*1*\dfrac{1}{3}\leq \dfrac{3}{16}.$

Hence $x = 3, y = 16 \Rightarrow x+y = 3+16=\boxed{19}.$