$(a+1)x^{2} +(b+1)y^{2} +(c+1)z^{2} \geq 2(xy+yz+zx)$ Prove $(a+1)x^{2} +(b+1)y^{2} +(c+1)z^{2} \geq 2(xy+yz+zx)$ for $a,b,c>0$ and $abc\geq1$.
My try: $(a+1)x^2-4xy+(b+1)y^2+(a+1)x^2-4xz+(c+1)z^2+(b+1)y^2-4yz+(c+1)z^2\geq0$
Then I had to prove that $(a+1)x^2-4xy+(b+1)y^2\geq2x^2-4xy+2y^2$, but I realized that there's no way of using $abc\geq1$.
Any help is appreciated!

We can assume at least two of ${x}$, ${y}$, ${z}$ non negative, otherwise we take their additive inverse. Let ${y}$, $z\ge0$.
Consider $x{}$ the only variable. The inequality can be regarded as a quadratic function of $x$. Since the coefficient of $x^2$is positive, we only need $\frac\Delta4\le0$, which requires
\[(a+1)\left((b+1)y^2+(c+1)z^2\right)\ge(y+z)^2+2(a+1)yz.\]Or $(ab+a+b)y^2+(ac+a+c)z^2\ge2(a+2)yz$. Using AM-GM, it suffices to show
\[(ab+a+b)(ac+a+c)\ge a+2.\]Expand the left side, since we have $ab\cdot ac\ge a$, $ab\cdot c$, $ac\cdot b\ge1$ and the rest $\ge0$, we're done.
