if $a,b,c$ are positive prove that $a^2+b^2+c^2+2abc+3\ge (1+a)(1+b)(1+c)$ if $a,b,c$ are positive prove that $$a^2+b^2+c^2+2abc+3\ge (1+a)(1+b)(1+c)$$

I have tried to use the  p-q-r method.
let $p=a+b+c,q=ab+bc+ca,r=abc$
we have to prove $$p^2-3q+r+2-p\ge 0$$
as $p^2\ge 3q$ we have to prove $$r+2-p\ge 0$$.
but i don't think this is true. Also i know the following inequalities but i can't figure out how to use them.

$pq\ge 9r$
$2p^3+9r\ge 7pq$
$q^3\ge 27r^2$ etc. etc.

Please note that i am looking for a solution only using pqr.I would be grateful if anyone can explain their logic behind the proof as it is the first time i am using 'pqr' method.
The inequality $r+2-p \geqslant 0$ does not true, example $a = 2, b = 1, c = \frac 12.$
We need to prove
$$p^2-p-3q+r+2 \geqslant 0,$$
equivalent to
$$(p^2+2r+1-4q)+(p^2-2p-2q+3) \geqslant 0.$$
Because
$$p^2-2p-2q+3 = (a-1)^2+(b-1)^2+(c-1)^2 \geqslant 0.$$
It's remain to prove that
$$p^2+2r+1 \geqslant 4q.$$
Using the AM-GM inequality, we have
$$2r+1 \geqslant 3\sqrt[3]{r^2} \geqslant \frac{9r}{p}.$$
Therefore, we will show that
$$p^2+ \frac{9r}{p}\geqslant 4q.$$
Which is Schur inequality.
Note. See here (page 4, Vietnamese).
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that:
$$a^2+b^2+c^2+abc+2\geq a+b+c+ab+ac+bc$$ or $f(v^2)\geq0,$ where
$$f(v^2)=9u^2-6v^2+w^3+2-(3u+3v^2)=9u^2-9v^2+w^3+2-3u.$$
But $f$ decreases, which says that it's enough to prove our inequality for a maximal value of $v^2$.
Now, $a$, $b$ and $c$ are roots of the equation $$(x-a)(x-b)(x-c)=0$$ or
$$x^3-3ux^2+3v^2x-w^3=0$$ or
$$-x^3+3ux^2+w^3=3v^2x.$$
Now, let $u$ and $w^3$ are constants and $v^2$ is changed. In this situation $a$, $b$ and $c$ are changed.
Since the last equation has three roots $a$, $b$ and $c$, we see that a graph of $g(x)=-x^3+3ux^2+w^3$ and the line $y=3v^2x$ have three common points and $v^2$ gets a maximal value for equality case of two numbers between $a$, $b$ and $c$.
Actually, if $a\leq b\leq c$ we can see that it happens for $b=c$.
Indeed, easy to draw a graph of $g$: $(0,w^3)$ it's a minimum point of $g$, $x_{max}=2u$ and the line $y=3v^2x$ goes across the origin.
If $v^2$ increases, so the line $y=3v^2x$ is  rotated and we can see, that $v^2$ may be maximal, if we want that our equation would have three positive roots, when $b=c$.
Id est, it's enough to prove our inequality for equality case of two variables.
Since our inequality is symmetric, it's enough to assume $b=c$ and we obtain a quadratic inequality of $a$ and the rest is smooth, which you can see here: show that $a^2+b^2+c^2+2abc+3\ge (1+a)(1+b)(1+c)$