Áp dụng BĐT Cô si, ta có :
\(a^4+b^4\ge2a^2b^2\)
\(b^4+c^4\ge2b^2c^2\)
\(c^4+a^4\ge2c^2a^2\)
\(\Rightarrow a^4+b^4+b^4+c^4+c^4+a^4\ge2a^2b^2+2b^2c^2+2c^2a^2\)
\(\Rightarrow a^4+b^4+c^4\ge a^2b^2+b^2c^2+c^2a^2\)( 1 )
Ta lại có :
\(a^2b^2+b^2c^2\ge2ab^2c\)
\(b^2c^2+c^2a^2\ge2bc^2a\)
\(c^2a^2+a^2b^2\ge2ca^2b\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2\ge ab^2c+bc^2a+ca^2b=abc\left(a+b+c\right)\)( 2 )
Từ ( 1 ) và ( 2 ) \(\Rightarrow a^4+b^4+c^4\ge abc\left(a+b+c\right)\forall a;b;c\)( Đpcm )
Ta có \(a^4+b^4+c^4\ge abc\left(a+b+c\right)\forall a;b;c>0\)
\(\Leftrightarrow a^4+b^4+c^4-a^2bc-b^2ac-c^2ab\ge0\)
\(\Leftrightarrow2a^4+2b^4+2c^4-2a^2bc-2b^2ac-2c^2ab\ge0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2+2a^2b^2+\left(b^2-c^2\right)^2+2b^2c^2+\left(c^2-a^2\right)^2-2a^2c^2-2b^2ac-2c^2ab\ge0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2+\left(b^2-c^2\right)^2-\left(c^2-a^2\right)^2+\left(a^2b^2+b^2c^2-2b^2ac\right)\)\(+\left(b^2c^2+c^2a^2-2c^2ab\right)+\left(a^2b^2+c^2a^2-2a^2bc\right)\ge0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2+\left(b^2-c^2\right)^2+\left(c^2-a^2\right)^2+\left(ab-bc\right)^2+\left(bc-ca\right)^2+\left(ab-ac\right)^2\ge0\)
Luôn đúng với mọi a,b,c
