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?
$$2(ab + bc + ca)-a^2 + b^2 + c^2>0$$
$$2ab + 2bc + 2ca-a^2 + b^2 + c^2>0$$
$$2b(a+c) + 2ca-a^2 + b^2 + c^2>0$$
...?

You can even prove more: if $a,b,c$ are the sides and $\alpha,\beta,\gamma$ the respectively opposite angles, summing up the three identities of the law of cosine, we get
$$
a^2+b^2+c^2=2ab\cos\gamma+2bc\cos\alpha+2ca\cos\beta
$$
which implies the statement, as all the cosines are less than $1$.
