a) \(a^2+b^2+c^2\ge ab+bc+ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\forall a,,b,c\in R\)
b)\(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)
ta có \(\left(a^2-b^2\right)^2+\left(b^2-c^2\right)^2+\left(a^2-c^2\right)^2\ge0\)
\(\Leftrightarrow a^4+b^4+c^4\ge a^2b^2+b^2c^2+c^2a^2\) (1)
ta cũng có \(\left(ab-bc\right)^2+\left(bc-ca\right)^2+\left(ca-ab\right)^2\ge0\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2\ge ab^2c+a^2bc+abc^2=abc\left(a+b+c\right)^{^{^{^{^{^{^{^{^{^{^{^{^{ }}}}}}}}}}}}}\) (2)
từ (1)(2) suy ra ĐPCM
