Find the minimum value of this

Let $a,b,c,d$ be nonnegative integers, such that $a+b+c+d=4$. Find the minimum value of $$\sum_{cyc}\frac{a}{b^3+4}.$$

My textbook says that the answer is $2/3$ achieved at $(2,2,0,0)$, but my method show that it is $1/17$ here it is

Since $a+b+c+d=4$, we have $a,b,c,d\le 4$ so $b^3+4\le 68$ hence $$\frac{a}{b^3+4} \ge\frac{a}{68}$$ Cycling we get the min is $1/17$, achieved at $(4,0,0,0)$, so where is the mistake?

Solution 1:

You have a mistake in replacement. Actually, the point $(a,b,c,d)=(4,0,0,0)$ leads to $$ \sum_{cyc}\frac{a}{b^3+4}=\frac{4}{0^3+4}+\frac{0}{0^3+4}+\frac{0}{0^3+4}+\frac{0}{4^3+4}=1. $$ What you found, is the value of $\sum_{cyc}\frac{a}{a^3+4}$.