Proof.
We'll prove an isolated fudging$$\frac{a+\sqrt{bc}}{\sqrt{(a+b)(a+c)}+\sqrt{bc}}\le \frac{2a+b+c}{2(a+b+c)}.\tag{1}$$
By AM-GM inequality,
\begin{align*}
&\frac{a+\sqrt{bc}}{\sqrt{(a+b)(a+c)}+\sqrt{bc}}\\&=\frac{(a+\sqrt{bc})(\sqrt{(a+b)(a+c)}-\sqrt{bc})}{a(a+b+c)}\\&=\frac{2(a+\sqrt{bc})\sqrt{(a+b)(a+c)}-2\sqrt{bc}(a+\sqrt{bc})}{2a(a+b+c)}\\&\le \frac{(a+\sqrt{bc})^2+(a^2+ab+bc+ca)-2\sqrt{bc}(a+\sqrt{bc})}{2a(a+b+c)}\\&=\frac{2a+b+c}{2(a+b+c)}.
\end{align*}
Take cyclic sum on $(1)$ we obtain the desired result. Equality holds iff $a=b=c>0.$
Denote the expression by $f(a, b, c)$.
Using $2uv \le u^2 + v^2$, we have
\begin{align*}
&a\sqrt{a^2+2bc}+b\sqrt{b^2+2ca}+c\sqrt{c^2+2ab}\\
\le{}& a\sqrt{a^2+b^2 + c^2}+b\sqrt{b^2+c^2 + a^2}+c\sqrt{c^2+a^2 + b^2}\\
={}& (a + b + c)\sqrt{a^2 + b^2 + c^2}. \tag{1}
\end{align*}
Also, by AM-GM, we have
$$(a + b + c)(ab + bc + ca) \ge 3\sqrt[3]{abc}\cdot 3\sqrt[3]{ab \cdot bc \cdot ca} = 9abc. \tag{2}$$
Using (1) and (2), letting $x = ab + bc + ca \in [0, 1/9]$ (using $(a + b + c)^2 \ge 3(ab + bc + ca)$), we have
\begin{align*}
f(a, b, c) &\le 3(a + b + c)(ab + bc + ca) + (a + b + c)\sqrt{a^2 + b^2 + c^2}\\
&= \sqrt 3\, (ab + bc + ca) + \frac{1}{\sqrt 3}
\sqrt{\frac13 - 2(ab + bc + ca)}\\
&= \sqrt 3\, x + \frac{1}{\sqrt 3}
\sqrt{\frac13 - 2x}\\
&\le \frac{2\sqrt 3}{9}
\end{align*}
where we use $a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca) = 1/3 - 2(ab + bc + ca)$, and
$$\left(\frac{2\sqrt 3}{9} - \sqrt 3\, x\right)^2
- \left(\frac{1}{\sqrt 3}
\sqrt{\frac13 - 2x}\right)^2 = \frac{1}{27}(9x - 1)^2 \ge 0.$$
Also, when $a = b = c = \frac{1}{3\sqrt 3}$,
we have $f(a, b, c) = \frac{2\sqrt 3}{9}$.
Thus, the maximum of $f(a,b,c)$ is $\frac{2\sqrt 3}{9}$.
Best Answer
By holder: $$\sum_{cyc}\sqrt[3]{\frac{1}{a}+\frac{2}{bc}+a+3b+c}\le \sqrt[3]{\sum_{cyc}({\frac{1}{a}+\frac{2}{bc}+a+3b+c})(1+1+1) \cdot (1+1+1)}$$ It hence suffices to prove $$\sqrt[3]{\sum_{cyc}({\frac{1}{a}+\frac{2}{bc}+a+3b+c})(1+1+1)(1+1+1)}\le \frac{6}{abc}$$ $$\sum_{cyc}({\frac{1}{a}+\frac{2}{bc}+a+3b+c})\le \frac{24}{a^3b^3c^3}\tag 1$$ Now let $p=a+b+c,q=ab+bc+ca=3,r=abc$ then we rewrite (1) as $$\frac{q}{r}+\frac{2p}{r}+5p\le \frac{24}{r^3}$$ $$\iff qr^2+2pr^2+5pr^3\le 24$$ Now as $q^2\ge 3pr \iff p\le \frac{3}{r}$ and $q=3$
it suffices to prove $$3r^2+6r+15r^2\le 24$$ which is true as $r\le 1$(by AM-GM)
Note in your original question you had $...a+2b+c$ i think its a typo and should be $a+3b+c$
all the same since $$\sum_{cyc}\sqrt[3]{({\frac{1}{a}+\frac{2}{bc}+a+2b+c})}\le \sqrt[3]{\sum_{cyc}({\frac{1}{a}+\frac{2}{bc}+a+3b+c})}$$ the same proof applies here here so we are done in either case