Positive numbers $x$, $y$ and $z$ satisfy $xy+yz+zx+xyz=4$, show that
\[x^2+y^2+z^2+3xyz\ge2(x+y+z).\]
I thought of $pqr$ method, let $\begin{cases}p=x+y+z,\\q=xy+yz+zx,\\r=xyz.\end{cases}$ so we have $q+r=4$, and needed to prove
\[p^2-2q+3r\ge2p.\]
Transform into $(p-1)^2-2q+3r\ge1$. Since $q$, $r$ have a condition, I wanted to get rid of $p$. Clearly $p>1$ (actually $p\ge3$), by $p^2\ge3q$, it suffices to prove
\[\left(\sqrt{3q}-1\right)^2-2q+3(4-q)\ge1.\]This resulted in $q+\sqrt{3q}\le6$, which is false because we can prove $q\ge3$. If we change $p$ into something else it will become only weaker so it cannot work. Perhaps we couldn't use $pqr$ method here.
Any method is okay, except for Lagrange Multipliers, of course.
Best Answer
pqr method:
It suffices to prove that $$p^2 - 2q + 3r - 2p + (4 - q - r) \ge 0$$ or $$p^2 - 2p + 4 + 2r - 3q\ge 0.$$
Degree three Schur's inequality yields $p^3 - 4pq + 9r \ge 0$ which results in $q \le \frac{p^3 + 9r}{4p}$.
It suffices to prove that $$p^2 - 2p + 4 + 2r - 3\cdot \frac{p^3 + 9r}{4p}\ge 0$$ or $$p(p-4)^2 - (27-8p)r \ge 0. \tag{1}$$
If $27 - 8p \le 0$, clearly (1) is true.
If $27 - 8p > 0$, using $p^3 \ge 27r$, we have $$p(p-4)^2 - (27-8p)r \ge p(p-4)^2 - (27-8p)\cdot \frac{p^3}{27} = \frac{8}{27}p(p+6)(p-3)^2 \ge 0.$$
We are done.