We have: \(ab\le\dfrac{\left(a+b\right)^2}{4}\Rightarrow\left(a+b\right)^2\ge4\Rightarrow a+b\ge2\left(a,b>0\right)\)

And \(a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)\ge ab\left(a+b\right)\ge2\)

By AM-GM's inequality we have:

\(\dfrac{a^3}{b+1}=a^3-\dfrac{a^3b}{b+1}\ge a^3-\dfrac{a^3b}{2\sqrt{b}}=a^3-\dfrac{a^3\sqrt{b}}{2}\)

Similar we have: \(\dfrac{b^3}{a+1}\ge b^3-\dfrac{b^3\sqrt{a}}{2}\)

\(\Rightarrow L.H.S\ge a^3+b^3-\dfrac{a^3\sqrt{b}+b^3\sqrt{a}}{2}\ge2-\dfrac{a^3\sqrt{b}+b^3\sqrt{a}}{2}\)

Need to prove: \(L.H.S\ge2-\dfrac{a^3\sqrt{b}+b^3\sqrt{a}}{2}\ge1=R.H.S\)

Use AM-GM's inequality:\(\sqrt{b}\le\dfrac{b+1}{2}\Rightarrow a^3\sqrt{b}\le\dfrac{a^3b+a^3}{2}\)

\(\Rightarrow\dfrac{a^3\sqrt{b}+b^3\sqrt{a}}{2}\le\dfrac{\dfrac{a^3b+a^3}{2}+\dfrac{ab^3+b^3}{2}}{2}=\dfrac{a^3b+ab^3+a^3+b^3}{4}\)

\(\Rightarrow2-\dfrac{a^3\sqrt{b}+b^3\sqrt{a}}{2}\ge2-\dfrac{a^3b+ab^3+a^3+b^3}{4}\)

Hence \(2-\dfrac{a^3b+ab^3+a^3+b^3}{4}\ge1\Leftrightarrow\dfrac{a^3b+ab^3+a^3+b^3}{4}\ge1\)

\(\Leftrightarrow a^3b+ab^3+a^3+b^3\ge4\)\(\Leftrightarrow a^3b+ab^3\ge2\) \(\left(a^3+b^3\ge2\right)\)

\(\Leftrightarrow\left(ab\right)^2\left(a+b\right)\ge2\). Right because \(ab=1;a+b\ge2\)

Done !!