$a,b,c ∈ \mathbb R^+$ and $a+b=1$ then find max value of $a^4b + ab^4$
So through AM-GM, I found : $$\frac{a+b}{2}\ge\sqrt{ab}\implies ab \le 1$$
Now I have, $$ab(a^3+b^3)\le\frac{1}{4}(a+b)(a^2+b^2-ab)$$ $$ab(a^3+b^3) \le \left(\frac{a^2+b^2}{4}-\frac{ab}{4}\right)$$
To me, it looks like I’m halfway there and just need to find something about $a^2 + b^2$ but haven’t found anything. Any type of help is welcomed.
Since $a,b ∈ R^+$ and $a+b=1$,
$$a^4b +ab^4=ab(a^3 + b^3) =$$ $$= ab[(a+b)^3-3ab(a+b)]=ab(1-3ab)=$$ $$=ab - 3a^2b^2=-3(a^2b^2-\frac{ab}{3})=$$ $$-3(a^2b^2-\frac{ab}{3}+\frac{1}{36})+\frac{1}{12}=$$ $$=-3(ab-\frac{1}{6})^2 +\frac{1}{12} \leq \frac{1}{12}$$