Giả sử \(2\left(a^4+b^4\right)\ge a^3b+ab^3+2a^2b^2\)
\(\Leftrightarrow2a^4+2b^4-a^3b-ab^3-2a^2b^2\ge0\)
\(\Leftrightarrow\left(a^4-a^3b\right)-\left(ab^3-b^4\right)+\left(a^4-2a^2b^2+b^4\right)\ge0\)
\(\Leftrightarrow a^3\left(a-b\right)-b^3\left(a-b\right)+\left(a^2-b^2\right)^2\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(a^3-b^3\right)+\left(a^2-b^2\right)^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a^2-ab+b^2\right)+\left(a^2-b^2\right)^2\ge0\) \(\forall a;b\) \(\left(1\right)\)
Lại có: \(a^2-ab+b^2=\left(a^2-2.a.\frac{b}{2}+\frac{b^2}{4}\right)+\frac{3b^2}{4}\)
\(=\left(a-\frac{b}{2}\right)^2+\frac{3b^2}{4}\ge0\) \(\forall a;b\) \(\left(2\right)\)
Từ (1) và (2) suy ra \(\left(a-b\right)^2\left(a^2-ab+b^2\right)+\left(a^2-b^2\right)^2\ge0\forall a;b\)
\(\Leftrightarrow2\left(a^4+b^4\right)\ge a^3b+ab^3+2a^2b^2\forall a;b\)
Vậy \(2\left(a^4+b^4\right)\ge a^3b+ab^3+2a^2b^2\) với mọi a;b
