Actually, this topic is very thought-provoking. The following is my answer. See if it's right. (Look at the picture, the text is latex code)

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.