Lemma. For positive $a$, $b$, $c$, $x$, $y$, $z$, it holds that $ax + by + cz + 2\sqrt{ab + bc + ca}\sqrt{xy + yz + zx} \le (a + b + c)(x + y + z)$

*Proof.* Use Cauchy-Schwarz inequality
$$\begin{aligned} ax + by + cz + 2\sqrt{ab + bc + ca}\sqrt{xy + yz + zx} & \le \sqrt{a^2 + b^2 + c^2}\sqrt{x^2 + y^2 + z^2} + \sqrt{ab + bc + ca}\sqrt{xy + yz + zx} + \sqrt{ab + bc + ca}\sqrt{xy + yz + zx} \\ & \le \sqrt{a^2 + b^2 + c^2 + 2ab + 2bc + 2ca}\sqrt{x^2 + y^2 + z^2 + 2xy + 2yz + 2zx} = (a + b + c)(x + y + z) \end{aligned}$$
Take $(x, y, z) = (\frac{a}{b+c}, \frac{b}{a+c}, \frac{c}{a+b})$ in the lemma. First we observe
$$\sqrt{xy + yz + zx} = \sqrt{\frac{a^2b + a^2c + b^2a + b^2c + c^2a + c^2b}{(a+b)(b+c)(c+a)}} = \frac{1}{\sqrt{1+2abc}}$$
By the lemma, we have
$$2\sqrt{ab + bc + ca}\sqrt{xy + yz + zx} \le (a + b + c)(x + y + z) - ax - by - cz$$
$$\frac{2\sqrt{ab + bc + ca}}{\sqrt{1 + 2abc}} \le ay + az + bx + bz + cx + cy$$
$$\frac{2\sqrt{ab + bc + ca}}{\sqrt{1 + 2abc}} \le x(b + c) + y(c + a) + z(a + b) = a + b + c$$
And finally, squaring both sides, we get the desired inequality
$$\frac{ab + bc + ca}{1 + 2abc} \le \frac{(a + b + c)^2}{4}$$