Question

[Thức Vương]

Cho a>0, b>0, c>0 thỏa mãn: \(\frac{a^3}{b+4c}\ge1\)

Tìm GTNN của: \(P=\frac{16a}{3}+\frac{1}{b}+\frac{1}{c}\)

Answer 1

[Phùng Minh Quân]

\(P=\frac{16a}{3}+\frac{1}{b}+\frac{4}{4c}\ge\frac{16a}{9}+\frac{16a}{9}+\frac{16a}{9}+\frac{9}{b+4c}\ge4\sqrt[4]{\frac{4096}{81}.\frac{a^3}{b+4c}}=\frac{32}{3}\)

"=" \(\Leftrightarrow\)\(\left(a;b;c\right)=\left(\frac{3}{2};\frac{9}{8};\frac{9}{16}\right)\)