Maximum Value of $a+b+c$

Question

Given that $a+\frac {11}b+\frac c4=20$, what is the maximum value of $a+b+c$? Here, $a,b,c$ are positive integers.

I tried to find the maximum from the AM-GM relationship but failed.

Answer 1

Notice

$$b > 0 \implies \frac{11}{b} > 0 \implies a + \frac{c}{4} < 20 \iff 4a+c < 80 \implies 4a + c \le 79$$

This leads to

$$b = \frac{11}{20- \left(a + \frac{c}{4}\right)} = \frac{44}{80 - (4a + c)} \le \frac{44}{80-79} = 44$$

Furthermore, $a \ge 1$ implies

$$a + c \le 79 - 3a \le 79 - 3 = 76$$

Combine these, we can conclude $a + b + c \le 76 + 44 = 120$.

Since this value $120$ is achieved by $(a,b,c) = (1,44,75)$ and $1 + \frac{11}{44} + \frac{75}{4} = 20$. The maximum value of $a + b + c$ equals to $120$.

Answer 2

Hint: $a=1$ and $\frac{44}{b}=76-c$ implies $b|44$.