Taking $ a=b=c=\frac{1}{3}$ and $ d=1$, we get $ k\le k_{max}$. Now lets prove this. Wlog $ a+b+c=1$. By AM-GM, $ 3^4(1+d)^5+2^4(1+2d)^5\ge 3^4(2\sqrt{d})^5+2^4(3\sqrt[3]{d^2})^5$ $ =6^4\cdot 5\cdot \left( \frac{2}{5}\cdot d^{\frac{5}{2}}+\frac{3}{5}\cdot d^{\frac{10}{3}}\right)\ge 6^4\cdot 5\cdot d^3\ge 6^4\cdot 5\cdot \frac{abc}{27} d^3$ So $ k=\frac{6^4\cdot 5}{27}$

Altheman wrote: Taking $ a=b=c=\frac{1}{3}$ and $ d=1$, we get $ k\le k_{max}$. Now lets prove this. Wlog $ a+b+c=1$. By AM-GM, $ 3^4(1+d)^5+2^4(1+2d)^5\ge 3^4(2\sqrt{d})^5+2^4(3\sqrt[3]{d^2})^5$ $ =6^4\cdot 5\cdot \left( \frac{2}{5}\cdot d^{\frac{5}{2}}+\frac{3}{5}\cdot d^{\frac{10}{3}}\right)\ge 6^4\cdot 5\cdot d^3\ge 6^4\cdot 5\cdot \frac{abc}{27} d^3$ So $ k=\frac{6^4\cdot 5}{27}$ Is this right??? at last step by AM-GM $ 6^4\cdot 5 \cdot\ 27 ?? $

Altheman wrote: Taking $ a=b=c=\frac{1}{3}$ and $ d=1$, we get $ k\le k_{max}$. Now lets prove this. Wlog $ a+b+c=1$. By AM-GM, $ 3^4(1+d)^5+2^4(1+2d)^5\ge 3^4(2\sqrt{d})^5+2^4(3\sqrt[3]{d^2})^5$ $ =6^4\cdot 5\cdot \left( \frac{2}{5}\cdot d^{\frac{5}{2}}+\frac{3}{5}\cdot d^{\frac{10}{3}}\right)\ge 6^4\cdot 5\cdot d^3\ge 6^4\cdot 5\cdot \frac{abc}{27} d^3$ So $ k=\frac{6^4\cdot 5}{27}$ I don't understand where this addressed that $\frac{6^4\cdot5}{27}$ is the maximum? Can you explain more Altheman?

The proof is composed into two parts - proving that $k_{max}$ is indeed the maximum, and proving that $k_{max}$ satisfies the inequality. The first line shows that by plugging $a=b=c=\frac{1}{3}$ and $d=1$, we get $k \le \frac{6^4 \cdot 5}{27}$. The second part shows that $k=\frac{6^4 \cdot 5}{27}$ satisfies the inequality. Therefore, we can conclude that $k_{max}$ is equal to $\frac{6^4 \cdot 5}{27}$, as desired.

How do you even approach these types of questions where you have to sort of "guess" the inequality you want to prove? Like, how would you know to plug in $(a,b,c,d)=(\frac 1 3, \frac 1 3, \frac 1 3, 1)$?