Prove that $a^2 + b^2 + c^2 \lt 2(ab + bc + ca)$ Given that $a, b$ and $c$ are the sides of a triangle.
How to prove that $a^2 + b^2 + c^2 \lt 2(ab + bc + ca)$?
Maybe any hint? Am I going to wrong direction?
$$2(ab + bc + ca)-a^2 + b^2 + c^2>0$$
$$2ab + 2bc + 2ca-a^2 + b^2 + c^2>0$$
$$2b(a+c) + 2ca-a^2 + b^2 + c^2>0$$
...?

Note that three sides $a,b,c$ in a triangle satisfy
$$ a<b+c, b<a+c, c<a+b. $$
So one has
$$ a^2<a(b+c), b^2<b(a+c), c^2<c(a+b). $$
Adding these three inequalities gives
$$ a^2+b^2+c^2<2ab+2bc+2ca. $$
