Because for any x, y, z? \ge? 0$, with $ (x+y+z) 2? \ge? 3(xy+yz+zx)$。
Substitute $ x = BC, y = CA, z = AB $ to get $ (BC+CA+AB) 2? \ge? 3abc(a+b+c)=9abc$, then $ab+bc+ca? \ge? Three dollars
So according to the average inequality,
\[\sqrt[3]{9abc(a^2+b^2+c^2)}=\sqrt[3]{3\sqrt{abc}? \cdot? 3\sqrt{abc}(a^2+b^2+c^2)}? \]
\[\le? \frac{3\sqrt{abc}+3\sqrt{abc}+a^2+b^2+c^2}{3}\le? \frac{2(ab+bc+ca)+a^2+b^2+c^2}{3}=3\].
It is proved that $ ABC (a 2+b 2+c 2) \ le? 3$. This is necessary inequality.