\(a^4+b^4\ge a^3b+ab^3\)

<=> \(a^4+b^4-a^3b-ab^3\ge0\)

<=> \(a^3\left(a-b\right)-b^3\left(a-b\right)\ge0\)

<=> \(\left(a^3-b^3\right)\left(a-b\right)\ge0\)

<=> \(\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)

Cause \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\forall a,b\in R\\a^2+ab+b^2=\left(a+\dfrac{b}{2}\right)^2+\dfrac{3b^2}{4}\ge0\end{matrix}\right.\)

So ...... 

a4+b4≥a3b+ab3

<=> a4+b4−a3b−ab3≥0

<=> a3(a−b)−b3(a−b)≥0

<=> (a3−b3)(a−b)≥0

<=> (a−b)2(a2+ab+b2)≥0

Cause ⎧⎩⎨⎪⎪⎪⎪(a−b)2≥0∀a,b∈Ra2+ab+b2=(a+b2)2+3b24≥0