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 :)

Since $x,y,z$ are sides of a triangle, we have 
$|x-y| < z $
Squaring both sides we get
$(x-y)^2=x^2 +y^2 -2xy < z^2$
Setting up similar inequalities and adding, we get the desired result.

Your proof doesn't work because
$(x+y+z)^2 \geq 0$ doesn't imply that 
$(x+y+z)^2 -4(xy+yz+zx) <0$, even when $x,y,z>0$
A simple counter-example is $(x,y,z)=(1,1,5)$
Obviously, this triple doesn't form a triangle, but contradicts the validity of the claim that $(x+y+z)^2 -4(xy+yz+zx) <0$, when $x,y,z>0$, which is the flaw in the given argument.
