$3\left(x^2 + y^2 +xy\right)\left(y^2 + z^2 +yz\right)\left(z^2 + x^2 +zx\right) \ge \left(x+y+z \right)^2\left(xy+yz+zx\right)^2$
My attempt :
For this I divided by $x^2y^2z^2$
We get
$3\left(\frac{x^2+y^2+xy}{xy}\right)\left(\frac{y^2+z^2+yz}{yz}\right)\left(\frac{x^2+z^2+xz}{xz}\right) \ge \left(x+y+z \right)^2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2 $
Now put
$ a = \frac{x}{y} + \frac{y}{x} + 1 $
$ b = \frac{y}{z} + \frac{z}{y} + 1 $
$ c = \frac{x}{z} + \frac{z}{x} + 1 $
So we get
$ 3abc \ge (a+b+c)^2 $
So we basically have to prove that
$ 3abc \ge (a+b+c)^2 $
I'am struck after this.
Could anyone help?
And yes $ x,y,z \in R $
Also do note $ a,b,c \ge 3 $ iff $x,y,z $ are of same sign