<s>[INST] 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?
$$2(ab + bc + ca)-a^2 + b^2 + c^2>0$$
$$2ab + 2bc + 2ca-a^2 + b^2 + c^2>0$$
$$2b(a+c) + 2ca-a^2 + b^2 + c^2>0$$
...?
 [/INST] Note that we have $$a^{2}>(b-c)^{2}\;,b^{2}>(a-c)^{2}\;,c^{2}>(b-a)^{2}$$
As $a,b,c$ are sides of a trinagle. Adding these, we get $$a^{2}+b^{2}+c^{2}>2a^{2}+2b^{2}+2c^{2}-2ab-2bc-2ca$$
Which is equivalent to $$a^2+b^2+c^2<2ab+2bc+2ca$$
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