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?

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 $$\sum_\text{cyc}{\sqrt\frac{a^2 + pbc}{b^2+c^2}} \geqslant u+\sqrt{\frac p{u^2-2}}\geqslant 2+\sqrt{\frac p2}\:.\quad\blacksquare$$

By AM-GM $$\sum_{cyc}\sqrt{\frac{a^2+pbc}{b^2+c^2}}\geq\sum_{cyc}\sqrt{\frac{a^2+\frac{2pb^2c^2}{b^2+c^2}}{b^2+c^2}}=\sum_{cyc}\frac{\sqrt{a^2b^2+a^2c^2+2pb^2c^2}}{b^2+c^2}$$ and it's enough to prove that: $$\sum_{cyc}\frac{\sqrt{ab+ac+2pbc}}{b+c}\geq2+\sqrt{\frac{p}{2}},$$ where $a$, $b$ and $c$ are non-negatives such that $ab+ac+bc\neq0$, or $$\sum_{cyc}\left(\frac{ab+ac+2pbc}{(b+c)^2}+\frac{2\sqrt{(bc+ac+2pab)(bc+ab+2pac)}}{(a+b)(a+c)}\right)\geq\left(2+\sqrt{\frac{p}{2}}\right)^2.$$ Now, by C-S $$\sqrt{(bc+ac+2pab)(bc+ab+2pac)}\geq bc+\sqrt{(ac+2pab)(ab+2pac)}=$$ $$=bc+a\sqrt{(c+2pb)(b+2pc)}\geq bc+a\left(\sqrt{2p}(b+c)+\frac{2bc(\sqrt{2p}-1)^2}{b+c}\right)$$ because $$(c+2pb)(b+2pc)-\left(\sqrt{2p}(b+c)+\frac{2bc(\sqrt{2p}-1)^2}{b+c}\right)^2=\frac{\left(\sqrt{2p}-1\right)^4bc(b-c)^2}{(b+c)^2}\geq0.$$ Id est, it's enough to prove that:

$$\sum_{cyc}\left(\frac{ab+ac+2pbc}{(b+c)^2}+\frac{2\left(bc+a\left(\sqrt{2p}(b+c)+\frac{2bc(\sqrt{2p}-1)^2}{b+c}\right)\right)}{(a+b)(a+c)}\right)\geq\left(2+\sqrt{\frac{p}{2}}\right)^2$$ or $$\sum_{cyc}\left(\tfrac{ab+ac+2pbc}{(b+c)^2}+\tfrac{2\left((\sqrt{2p}+1)(a^2b+a^2c)+2(2p-\sqrt{2p}+1)abc\right)}{(a+b)(a+c)(b+c)}\right)\geq\left(2+\sqrt{\frac{p}{2}}\right)^2.$$ Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, we need to prove that: $$\sum_{cyc}\left(\frac{a}{b+c}+\tfrac{2pbc}{(b+c)^2}+\tfrac{2\left((\sqrt{2p}+1)(a^2b+a^2c)+2(2p-\sqrt{2p}+1)abc\right)}{(a+b)(a+c)(b+c)}\right)\geq\left(2+\sqrt{\frac{p}{2}}\right)^2$$ or $$\frac{\sum\limits_{cyc}a(a+b)(a+c)}{\prod\limits_{cyc}(a+b)}+\frac{2p\sum\limits_{cyc}bc(a+b)^2(a+c)^2}{\prod\limits_{cyc}(a+b)^2}+$$ $$+\sum_{cyc}\frac{2\left((\sqrt{2p}+1)(a^2b+a^2c)+2(2p-\sqrt{2p}+1)abc\right)}{(a+b)(a+c)(b+c)}\geq\left(2+\sqrt{\frac{p}{2}}\right)^2$$ or $$\frac{27u^3-18uv^2+3w^3}{9uv^2-w^3}+\frac{2p\sum\limits_{cyc}bc(a^2+3v^2)^2}{(9uv^2-w^3)^2}+$$ $$+\frac{2\left((\sqrt{2p}+1)(9uv^2-3w^3)+6(2p-\sqrt{2p}+1)w^3\right)}{9uv^2-w^3}\geq\left(2+\sqrt{\frac{p}{2}}\right)^2,$$ which says that our inequality is equivalent to $f(w)^3\geq0,$ where $$f(w^3)=kw^6+A(u,v^2)w^3+B(u,v^2)$$ and $$k=-3+6p+6(\sqrt{2p}+1)-12(2p-\sqrt{2p}+1)-\left(2+\sqrt{\frac{p}{2}}\right)^2=$$ $$=-\frac{1}{2}(37p-32\sqrt{2p}+26)<0.$$ We see that $f$ is a concave function, which says that $f$ gets a minimal value for an extreme value of $w^3$, which by $uvw$ (about $uvw$ see here: https://artofproblemsolving.com/community/c6h278791 ) happens in the following cases.

1. $w^3=0$.

Let $c=0$.

We obtain: $$\sum_{cyc}\left(\frac{ab+ac+2pbc}{(b+c)^2}+\tfrac{2\left((\sqrt{2p}+1)(a^2b+a^2c)+2(2p-\sqrt{2p}+1)abc\right)}{(a+b)(a+c)(b+c)}\right)\geq\left(2+\sqrt{\frac{p}{2}}\right)^2$$ it's $$\frac{a}{b}+\frac{b}{a}+\frac{2pab}{(a+b)^2}\geq2+\frac{p}{2}$$ or $$\frac{1}{ab}\geq\frac{p}{2(a+b)^2},$$ which is true because by AM-GM $$\frac{2(a+b)^2}{ab}\geq8>p.$$ 2. Two variables are equal.

Let $b=c=1$.

In this case $$\sum_{cyc}\left(\frac{ab+ac+2pbc}{(b+c)^2}+\tfrac{2\left((\sqrt{2p}+1)(a^2b+a^2c)+2(2p-\sqrt{2p}+1)abc\right)}{(a+b)(a+c)(b+c)}\right)\geq\left(2+\sqrt{\frac{p}{2}}\right)^2$$ it's $$a((a-1)^2+(2\sqrt{2p}-1)^2)\geq0$$ and we are done!