The inequality $\,2+\sqrt{\frac p2}\leq\sum\limits_\text{cyc}\sqrt{\frac{a^2+pbc}{b^2+c^2}}\,$ where $0\leq p\leq 2$ is: Probably true! Provably true? Let $p$ be a positive parameter in the range from $0$ to $2$.

Can one prove that
$$2 +\sqrt{\frac p2} \;\leqslant\;\sqrt{\frac{a^2 + pbc}{b^2+c^2}}
 \,+\,\sqrt{\frac{b^2 +pca}{c^2+a^2}}\,+\,\sqrt{\frac{c^2 +pab}{a^2+b^2}}\quad?\tag{1}$$
Where $\,a,b,c\in\mathbb R^{\geqslant 0}\,$ and at most one variable equals zero.

The inequality $(1)$ is homogeneous of degree zero with regard to $a,b,c$.
Equality occurs if two variables coincide and the third one is zero.
To provide some plausibility to $(1)$ the two boundary cases $\,p=2\,$ and $\,p=0\,$ are
proved:
$p=2$
is the harder bit.
W.l.o.g. assume $\,a\geqslant b\geqslant c\,$ and $\,a,b>0$. Let
$u=\sqrt{\frac ab}\,+\,\sqrt{\frac ba}$, then $2\leqslant u$, and $u=2$ iff $a=b$.
I)$\:$ Let's show that
$$u\:\leqslant\:\sqrt{\frac{a^2 + bc}{b^2+c^2}} \,+\,\sqrt{\frac{b^2 +ac}{a^2+c^2}}\:.\tag{2}$$
The following expression is positive:
$$\begin{align}
 & \frac{a^2+bc}{b^2+c^2} -\frac ab & +\quad &\frac{b^2+ac}{a^2+c^2} -\frac ba\\[2ex]
=\;\; & \frac{ab(a-b)+b^2c-ac^2}{b(b^2+c^2)} & +\quad &\frac{-ab(a-b)+a^2c-bc^2}{a(a^2+c^2)}\tag{3}\\[2ex]
\geqslant\;\;  & \frac{ac(a-c)+bc(b-c)}{a(a^2+c^2)} \;\geqslant 0
\end{align}$$
The first summand in $(3)$ has been diminished by increasing the denominator, while
its numerator $\,ab(a-b) +b^2c -ac^2 =(a-b)(b-c)(a-c) + c(a-c)^2 + c^2(b-c)\,$ cannot get negative.
$(2)$ now follows from
$$\begin{split}
u^2\:=\:\frac ab + 2 +\frac ba \: & \leqslant\:\frac{a^2+bc}{b^2+c^2}
+ 2\,\underbrace{\sqrt{\frac{a^2+bc}{a^2+c^2}}}_{\geqslant 1}\;\underbrace{\sqrt{\frac{b^2+ac}{b^2+c^2}}}_{\geqslant 1}+ \frac{b^2+ac}{a^2+c^2}\\[2ex]
& =\:\left(\sqrt{\frac{a^2 + bc}{b^2+c^2}} +\sqrt{\frac{b^2 +ac}{a^2+c^2}}\:\right)^2
\end{split}$$
II)$\:$ The remaining square root summand in $(1)$ is also bounded below in terms of $u$ since one has
$$\frac 1{u^2-2} \:=\:\frac{ab}{a^2+b^2}\quad\implies\quad
\sqrt{\frac 2{u^2-2}} \:\leqslant\: \sqrt{\frac{c^2 +2ab}{a^2+b^2}}$$
III)$\:$ Applying $3$-AGM finally proves $(1)$:
$$\begin{split}\sum_\text{cyc}{\sqrt\frac{a^2 + 2bc}{b^2+c^2}} \;\geqslant\;
u+\sqrt{\frac 2{u^2-2}} &\:=\:\sqrt{\frac{u^2}4} +\sqrt{\frac{u^2}4} +\sqrt{\frac 2{u^2-2}}\\[2ex]
&\:\geqslant\:3\sqrt{\left(\frac{u^4}{8(u^2-2)}\right)^{1/3}} \:\geqslant\:3\end{split}$$

$p=0$
is more relaxing.
Only $2$-AGM in the form $\,a\sqrt{b^2+c^2}\leqslant\frac12\left(a^2+b^2+c^2\right)$ is needed:
$$\frac a{\sqrt{b^2+c^2}} + \frac b{\sqrt{c^2+a^2}}+ \frac c{\sqrt{a^2+b^2}}
\;=\;\sum_\text{cyc}\frac{a^2}{a\sqrt{b^2+c^2}}
\;\geqslant\;\sum_\text{cyc}\frac{2a^2}{a^2+b^2+c^2} \;=\;2$$

$0<p<2$
returns to the question.
With just some ideas how to catch the "remaining" $p$-values:

*

*The above method for $p=2$ may possibly be stretched down until the $p=1$ instance:
$$2 +\frac{\sqrt 2}{2} \;\leqslant\;\sum_\text{cyc}{\sqrt\frac{a^2+bc}{b^2+c^2}}$$
This has been detailed by mathlove in his answer.

*Interpolation with regard to $p$ (more a buzz word than substantial ...)

*A concavity argument as the two end points $p=0$ and $p=2$ are known: Could proving the second derivative with respect to $p$ being negative path a way towards a proof?

 A: This is a partial answer.
This answer proves that for $1\leqslant p\leqslant 2$, the inequality $(1)$ holds.
Proof :
(You have already proved that for $p=2$, $(1)$ holds. The idea used in the case $p=2$ works for the case $1\leqslant p\leqslant 2$.)
W.l.o.g. assume $\,a\geqslant b\geqslant c\,$ and $\,a,b>0$. Let
$u=\sqrt{\frac ab}\,+\,\sqrt{\frac ba}$, then $2\leqslant u$, and $u=2$ iff $a=b$.
You have already proved that
$$u\:\leqslant\:\sqrt{\frac{a^2 + bc}{b^2+c^2}} \,+\,\sqrt{\frac{b^2 +ac}{a^2+c^2}}\:.\tag{2}$$
Since
$$\sqrt{\frac{c^2 +pab}{a^2+b^2}}\geqslant\sqrt{\frac{pab}{a^2+b^2}}=\sqrt{\frac{p}{\frac ab+\frac ba}}=\sqrt{\frac{p}{u^2-2}}$$
one has, using $(2)$,
$$\sum_\text{cyc}{\sqrt\frac{a^2 + pbc}{b^2+c^2}} \geqslant
u+\sqrt{\frac p{u^2-2}}$$
Here, one can have
$$u+\sqrt{\frac p{u^2-2}}\geqslant 2+\sqrt{\frac p2}$$
since
$$\begin{align}&\bigg(u+\sqrt{\frac p{u^2-2}}\bigg)-\bigg(2+\sqrt{\frac p2}\bigg)
\\\\&=\frac{(u-2)\sqrt{2(u^2-2)}+\sqrt p(\sqrt 2-\sqrt{u^2-2})}{\sqrt{2(u^2-2)}}
\\\\&=\frac{(u-2)\sqrt{2(u^2-2)}+\sqrt p\cdot\dfrac{(2-u)(2+u)}{\sqrt 2+\sqrt{u^2-2}}}{\sqrt{2(u^2-2)}}
\\\\&=\frac{u-2}{\sqrt{2(u^2-2)}}\bigg(\sqrt{2(u^2-2)}-\dfrac{\sqrt p(2+u)}{\sqrt 2+\sqrt{u^2-2}}\bigg)
\\\\&=\frac{(u-2)(u+2)}{(\sqrt 2+\sqrt{u^2-2})\sqrt{2(u^2-2)}}\bigg(\frac{2\sqrt{u^2-2}+(u^2-2)\sqrt 2}{u+2}-\sqrt p\bigg)
\\\\&=\underbrace{\frac{(u-2)(u+2)}{(\sqrt 2+\sqrt{u^2-2})\sqrt{2(u^2-2)}}}_{\text{non-negative}}\bigg(\underbrace{\frac{2\sqrt{1-\frac{2}{u^2}}+(u-\frac 2u)\sqrt 2}{1+\frac 2u}}_{\text{increasing}}-\sqrt p\bigg)
\\\\&\geqslant\frac{(u-2)(u+2)}{(\sqrt 2+\sqrt{u^2-2})\sqrt{2(u^2-2)}}\bigg(\frac{2\sqrt{1-\frac{2}{2^2}}+(2-\frac 22)\sqrt 2}{1+\frac 22}-\sqrt p\bigg)
\\\\&=\frac{(u-2)(u+2)}{(\sqrt 2+\sqrt{u^2-2})\sqrt{2(u^2-2)}}\bigg(\sqrt 2-\sqrt p\bigg)
\\\\&\geqslant 0\:.\end{align}$$
Therefore, it follows that
$$\sum_\text{cyc}{\sqrt\frac{a^2 + pbc}{b^2+c^2}} \geqslant
u+\sqrt{\frac p{u^2-2}}\geqslant 2+\sqrt{\frac p2}\:.\quad\blacksquare$$
