Lời giải:
Ta có: Theo BĐT Cô-si thì:
\(4=2^a+4^b+8^c=2^a+2^{2b}+2^{3c}\)
\(\geq 3\sqrt[3]{2^a.2^{2b}.2^{3c}}=3\sqrt[3]{2^{a+2b+3c}}\)
\(\Rightarrow a+2b+3c\leq 3\log_2(\frac{4}{3})\)
\(\frac{a}{6}+\frac{b}{3}+\frac{c}{2}=\frac{a+2b+3c}{6}\leq \frac{\log_2(\frac{4}{3})}{2}\)
hay \(m=(\frac{a}{6}+\frac{b}{3}+\frac{c}{2})_{\max}=\frac{\log_2(\frac{4}{3})}{2}\)