If $x, y, z$ are the side lengths of a triangle, prove that $x^2 + y^2 + z^2 < 2(xy + yz + xz)$ 

Question: If $x, y, z$ are the side lengths of a triangle, prove that $x^2 + y^2 + z^2 < 2(xy + yz + xz)$



My solution: Consider
$$x^2 + y^2 + z^2 < 2(xy + yz + xz)$$
Notice that $x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)$
Hence
$$(x+y+z)^2-2(xy+yz+xz) < 2(xy + yz + xz)$$
$$ (x+y+z)^2 - 4(xy+yz+xz) < 0 $$
As $x,y,z$ are all greater than zero as a side length of a triangle can not be negative. And because $(x+y+z)^2 >0$ for all real $x,y,z$ therefore the whole expression is less than zero. $Q.E.D$
Am I correct? Or could I be more "rigorous" I am a highschool student and getting into proof so any tips would be appreciated as well :)

$x,y$ and $z$ are sides of a triangle. So, $y+z\gt x,\;  x+y\gt z,\;  z+x\gt y$. 
Now you have to prove that $$x^2+y^2+z^2\lt2\left(\sum xy\right).$$
Just a simple look through the right hand term.
$$2\left(\sum xy\right)=2xy+2yz+2zx\\=x\color{blue}{(y+z)}+y\color{blue}{(z+x)}+z\color{blue}{(x+y)}\gt x\times \color{blue}{x}+y\times \color{blue}{y}+z\times \color{blue}{z}=x^2+y^2+z^2$$
So, now, you derived $$x^2+y^2+z^2\lt2\left(\sum xy\right).$$
