Can You Solve This Challenging Inequality Problem?

[anemone]

Given positive real numbers $a,\,b,\,c$ and $d$ that satisfy the following inequalities:

$a \le 1 \\a+4b \le 17\\a+4b+16c \le273\\a+4b+16c+64d \le4369$

Find the minimum value of $\dfrac{1}{d}+\dfrac{2}{4c+d}+\dfrac{3}{16b+4c+d}+\dfrac{4}{64a+16b+4c+d}$.

 

[lfdahl]

Spoiler: My suggested solution

The coefficients in the denominators by necessity gives the following order of priority: $a,b,c,d$, i.e. $a$ must be our first choice and $a$ must necessarily be chosen as large as possible: $a = 1$, in order to maximize the last term in the given expression. The next to be prioritized is $b$ with the help of the second inequality, which implies: $b = 4$. Next in the priority list is $c$ and from the third inequality criterion, we have: $c = 16$. At last comes $d$ in the last inequality: $d = 64$.

Thus our minimum value must be: $\frac{4}{64} = \frac{1}{16}$.