Prove that $(a+b-c)^2+(b+c-a)^2+(c+a-b)^2>ab+bc+ca$ How can I prove $(a+b-c)^2+(b+c-a)^2+(c+a-b)^2>ab+bc+ca$? 
We have  
$(a+b-c)^2+=a^2+b^2+b^2+2ab-2bc-2ca$,
$(b+c-a)^2+=b^2+c^2+a^2+2bc-2ca-2ab$
$(c+a-b)^2+=c^2+a^2+b^2+2ca-2ab-2bc$
But I don't know how to show the required result?
Please give hint. Thank you 

We have
$$(a+b-c)^2+(b+c-a)^2+(c+a-b)^2 = 3a^2+3b^2+3c^2-2ab-2bc-2ca$$
From AM-GM, we have $a^2+b^2 \geq 2ab$, $b^2+c^2 \geq 2bc$ and $c^2+a^2 \geq 2ca$. Hence,
$$2(a^2+b^2+c^2) \geq 2(ab+bc+ca) \implies 3(a^2+b^2+c^2) \geq 3ab+3bc+3ca$$
Hence,
\begin{align}
(a+b-c)^2+(b+c-a)^2+(c+a-b)^2 & = 3a^2+3b^2+3c^2-2ab-2bc-2ca\\
& \geq 3ab+3bc+3ca-2ab-2bc-2ca\\
& = ab+bc+ca
\end{align}
which is what you want.
