a.
\(a^2b^2+b^2c^2+c^2a^2\ge abc\left(a+b+c\right)\)
\(\Leftrightarrow2a^2b^2+2b^2c^2+2c^2a^2\ge2abc\left(a+b+c\right)\)
\(\Leftrightarrow\left(a^2b^2-2a^2bc+a^2c^2\right)+\left(a^2b^2-2ab^2c+b^2c^2\right)+\left(b^2c^2-2abc^2+c^2a^2\right)\ge0\)
\(\Leftrightarrow\left(ab-ac\right)^2+\left(ab-bc\right)^2+\left(bc-ca\right)^2\ge0\) (luôn đúng)
Vậy BĐT ban đầu đúng
Dấu "=" xảy ra khi \(a=b=c\)
b.
Từ câu a ta có:
\(a^2b^2+b^2c^2+c^2a^2\ge abc\left(a+b+c\right)\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\ge3abc\left(a+b+c\right)\)
\(\Leftrightarrow\left(ab+bc+ca\right)^2\ge3abc\left(a+b+c\right)\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
