Inequality: $2(p^2+q^2+r^2)+2(pq+qr+rp)\ge pqr$ I need to determine the range of $p,q,r$ such that $2(p^2+q^2+r^2)+2(pq+qr+rp)\ge pqr$. I am not given any other information except that $p,q,r\in \mathbb{R}$. I haven't solved a problem like this before, so I'm really stuck. I'm thinking that it might be true for all real numbers $p,q,r$ but then I would need a proof but I couldn't find one.
Thanks!
EDIT: Sorry, I forgot to add:
I wrote it as $(p+q)^2+(q+r)^2+(r+p)^2\ge pqr$.
 A: This doesn't hold true in general.
For example, let $p=q=r$, you will get $$12p^2\geq p^3.$$ You can see it is not true when $p>12$.
A: Let me prove it for all $p,q,r \in (-1,\sqrt{3}-1)$.
First, if one or three of $p,q,r$ are nonpositive, it is clear the inequality holds.
Now consider $p,q,r$ are all positive. Then $|r| < 4$. Since for any $p,q$, $(p+q)^2\geq 4pq$.
It is clear that $$(p+q)^2+(p+r)^2+(q+r)^2\geq (p+q)^2\geq 4pq \geq pqr.$$
Next consider two of them are nonpositive. Without loss of generality, let $p\leq 0$ and $q\leq 0$. Again, the above argument holds. 

The whole idea is using $(p+q)^2\geq 4pq$ which holds for any $p,q\in\mathbb{R}$.
A: may be almagest 's suggestion is more general. 
$ 2r^2+(2p+2q-pq)r+2(p^2+q^2+pq)\ge0$ will hold when 
$(2p+2q-pq)^2-4*2*2(p^2+q^2+pq)\le 0 \iff (p^2-4p-12)q^2+(-4p^2-8p)q-12p^2 \le 0 \iff (p^2-4p-12)<0 \cap  (-4p^2-8p)^2+4*12p^2(p^2-4p-12) \le0 \iff -2<p<6 \cap -2 \le p \le4 \implies -2<p\le4$.
but when $p=-2$ it is also true for $\Delta_r <0 $.
so one of $p,q,r$ fall in $[-2,4]$, the inequality will hold. the other two have no limitation.
