For positive real numbers $x$, $y$, $z$, from the arithmetic-geometric-mean inequality,
$$2x + y + z = (x + y) + (x + z) \geq 2 \sqrt{(x + y)(x + z)}$$
we obtain
$$\frac{1}{(2x + y + z)^2} \leq \frac{1}{4(x + y)(x + z)}$$
Applying this to the left-hand side terms of the inequality to prove, we get
$$\begin{align*} \frac{1}{(2a + b + c)^2} &+ \frac{1}{(2b + c + a)^2} + \frac{1}{(2c + a + b)^2} \\ &\leq \frac{1}{4(a + b)(a + c)} + \frac{1}{4(b + c)(b + a)} + \frac{1}{4(c + a)(c + b)} \\ &= \frac{(b + c) + (c + a) + (a + b)}{4(a + b)(b + c)(c + a)} = \frac{a + b + c}{2(a + b)(b + c)(c + a)} \tag{1} \end{align*}$$
A second application of the inequality of the arithmetic-geometric mean yields
$$a^2 b + a^2 c + b^2 a + b^2 c + c^2 a + c^2 b \geq 6abc$$
or, equivalently,
$$\begin{equation*} 9(a + b)(b + c)(c + a) \geq 8(a + b + c)(ab + bc + ca) \tag{2} \end{equation*}$$
The supposition $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = a + b + c$ can be written as
$$\begin{equation*} ab + bc + ca = abc(a + b + c) . \tag{3} \end{equation*}$$
Applying the arithmetic-geometric-mean inequality $x^2 y^2 + x^2 z^2 \geq 2x^2 y z$ thrice, we get
$$a^2 b^2 + b^2 c^2 + c^2 a^2 \geq a^2 b c + a b^2 c + a b c^2$$
which is equivalent to
$$\begin{equation*} (ab + bc + ca)^2 \geq 3abc(a + b + c) \tag{4} \end{equation*}$$
Combining (1), (2), (3), and (4), we will finish the proof:
$$\begin{aligned} \frac{a + b + c}{2(a + b)(b + c)(c + a)} &= \frac{(a + b + c)(ab + bc + ca)}{2(a + b)(b + c)(c + a)} \cdot \frac{ab + bc + ca}{abc(a + b + c)} \cdot \frac{abc(a + b + c)}{(ab + bc + ca)^2} \\ &\leq \frac{9}{2 \cdot 8} \cdot 1 \cdot \frac{1}{3} = \frac{3}{16} \end{aligned}$$

Equivalently, we prove the homogenized inequality
$$\frac{(a + b + c)^2}{(2a + b + c)^2} + \frac{(a + b + c)^2}{(a + 2b + c)^2} + \frac{(a + b + c)^2}{(a + b + 2c)^2} \leq \frac{3}{16}(a + b + c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)$$
for all positive real numbers $a$, $b$, $c$. Without loss of generality we choose $a + b + c = 1$. Thus, the problem is equivalent to prove for all $a$, $b$, $c > 0$, fulfilling this condition, the inequality
$$\begin{equation*} \frac{1}{(1 + a)^2} + \frac{1}{(1 + b)^2} + \frac{1}{(1 + c)^2} \leq \frac{3}{16}\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \tag{5} \end{equation*}$$
Applying Jensen's inequality to the function $f(x) = \frac{x}{(1 + x)^2}$, which is concave for $0 \leq x \leq 2$ and increasing for $0 \leq x \leq 1$, we obtain
$$\alpha \frac{a}{(1 + a)^2} + \beta \frac{b}{(1 + b)^2} + \gamma \frac{c}{(1 + c)^2} \leq (\alpha + \beta + \gamma) \frac{A}{(1 + A)^2}, \quad \text{ where } \quad A = \frac{\alpha a + \beta b + \gamma c}{\alpha + \beta + \gamma} .$$
Choosing $\alpha = \frac{1}{a}$, $\beta = \frac{1}{b}$, and $\gamma = \frac{1}{c}$, we can apply the harmonic-arithmetic-mean inequality
$$A = \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \leq \frac{a + b + c}{3} = \frac{1}{3} < 1$$
Finally we prove (5):
$$$$
$$\begin{aligned} \frac{1}{(1 + a)^2} + \frac{1}{(1 + b)^2} + \frac{1}{(1 + c)^2} &\leq \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \frac{A}{(1 + A)^2} \\ &\leq \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \frac{\frac{1}{3}}{\left(1 + \frac{1}{3}\right)^2} = \frac{3}{16}\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \end{aligned}$$