For positive real $a,b$ Prove that $\frac{a^3b}{(a+b)^4}\le \frac{27}{256}$

Question

For positive real $a,b$ Prove that $\frac{a^3b}{(a+b)^4}\le \frac{27}{256}$

I tried using weighted AM-GM inequality with weight 3 with $a$ and weight 1 with $b$ $$\frac{3a+b}{4} \ge (a^3b)^{1/4}$$ $$\frac{(3a+b)^4}{256} \ge a^3b$$ but I don't know how to relate $3a+b$ with $a+b$

Also solve using AM-GM-HM and weighted AM-GM-HM only because the book hasn't introduced any other inequalities at this point.

Answer 1

Because by AM-GM $$\frac{a^3b}{(a+b)^4}=\frac{a^3b}{\left(3\cdot\frac{a}{3}+b\right)^4}\leq\frac{a^3b}{\left(4\sqrt[4]{\left(\frac{a}{3}\right)^3b}\right)^4}=\frac{27}{256}.$$