Question

Cho a, b, c > 0 và \(a+\sqrt{ab}+\sqrt[3]{abc}=\dfrac{4}{3}\) .

Tìm MIN : \(Q=a+b+c\)

Answer 1

\(\dfrac{4}{3}\le a+\sqrt{ab}+\sqrt[3]{abc}=a+\sqrt[]{\dfrac{a}{2}.2b}+\sqrt[3]{\dfrac{a}{4}.b.4c}\)

\(\le a+\dfrac{1}{2}\left(\dfrac{a}{2}+2b\right)+\dfrac{1}{3}\left(\dfrac{a}{4}+b+4c\right)=\dfrac{4}{3}\left(a+b+c\right)\)

\(\Rightarrow Q\ge1\)

\(Q_{min}=1\) khi \(\left(a;b;c\right)=\left(\dfrac{16}{21};\dfrac{4}{21};\dfrac{1}{21}\right)\)