\(\dfrac{a^3+b^3}{2}\ge\left(\dfrac{a+b}{3}\right)^3\)

<=> \(8.\dfrac{a^3+b^3}{2}\ge a^3+3a^2b+3ab^2+b^3\)

<=> \(3\left(a^3+b^3\right)\ge3a^2b+3ab^2\)

<=> \(a^3+b^3-a^2b-ab^3\ge0\)

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

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

So ......