\(P=\sqrt{3a^2+2ab+3b^2}+...+\sqrt{3c^2+2ac+3a^2}\)
ta có:
\(\sqrt{3a^2+2ab+3b^2}\ge\sqrt{2}\left(a+b\right)\Leftrightarrow3a^2+2ab+3b^2\ge2a^2+4ab+2b^2\Leftrightarrow\left(a-b\right)^2\ge0\left(đ\right)\)
\(\text{tương tự suy ra:}P\ge2\sqrt{2}\left(a+b+c\right)\)