Looks tough! [part 2]

I'll certainly generalise this :-).
Let $a+b+c+d=\mathfrak{M}$ and we have to find maximum value of $\mathfrak{K}=a^{p}b^{q}c^{r}d^{s}$ .

Writing $a=\frac{a}{p}+\frac{a}{p}\cdots\text{p times},\\ b=\frac{b}{q}+\frac{b}{q}\cdots\text{q times},\\ c=\frac{c}{r}+\frac{c}{r}\cdots\text{r times},\\d=\frac{d}{s}+\frac{d}{s}\cdots\text{s times}\\$ and applying AM-GM. $\Large(\dfrac{\mathfrak{K}}{p^pq^qr^rs^s})^{\frac{1}{p+q+r+s}}\leq\dfrac{\mathfrak{M}}{p+q+r+s}$ $\Large \Rightarrow \mathfrak{K}\leq(\dfrac{\mathfrak{M}}{p+q+r+s})^{p+q+r+s}\times p^pq^qr^rs^s$ For equality $\color{#3D99F6}{\dfrac{a}{p}=\dfrac{b}{q}=\dfrac{c}{r}=\dfrac{d}{s}=\dfrac{\mathfrak{M}}{p+q+r+s}}$ .
In this question $\color{#D61F06}{p=1,q=2,r=3,s=4}$ and $\color{#D61F06}{\mathfrak{M}=20}$ . Substituting values we get : $\huge \mathfrak{K}\leq \color{#007fff}{2^{20}\times 3^3}$ (In this question, for equality $\color{#3D99F6}{ a=\frac{b}{2}=\frac{c}{3}=\frac{d}{4}=2}$ )