a) BĐT $\Leftrightarrow
a^2+b^2+c^2+a^2b^2+b^2c^2+c^2a^2-6abc\geq 0$
$\Leftrightarrow (a-bc)^2+(b-ca)^2+(c-ab)^2\geq
0$, đúng
b) Ta có:
\(
\frac{a^{2}}{3}+b^{2}+c^{2}>ab+bc+ca
\)
\(
\Leftrightarrow \frac{a^{2}}{3}+b^{2}+c^{2}-ab-bc-ca>0
\)
\(
\Leftrightarrow [\frac{a^{2}}{4}+\left ( b+c \right )^{2}-a\left ( b+c \right )]+\frac{a^{2}}{12}-3bc>0
\)
\(
\Leftrightarrow \left ( \frac{a}{2}-b-c \right )^{2}+\frac{a^{2}-36bc}{12}>0
\) (do \(
abc=1
\))
\(
\Leftrightarrow \left ( \frac{a}{2}-b-c \right )^{2}+\frac{a^{3}-36}{12a}>0:
\) Đúng vì \(
a>\sqrt[3]{36}=> a>0, a^{3}>36.
\)