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

Alternatively, by cosine law
\begin{align*}
  b^2+c^2-a^2 &= 2bc\cos A \\
  c^2+a^2-b^2 &= 2ca\cos B \\
  a^2+b^2-c^2 &= 2ab\cos C \\
  a^2+b^2+c^2 &= 2(bc\cos A+ca\cos B+ab\cos C) \\
  & \le 2(bc+ca+ab)
\end{align*}
