Question

tìm max s= \(\dfrac{\sqrt[3]{\left(a-2\right)\left(b-3\right)}}{a+b}\) với \(\left\{{}\begin{matrix}a\ge2\\b\ge3\end{matrix}\right.\)

Answer 1

\(S=\dfrac{\sqrt[3]{\left(a-2\right)\left(b-3\right)}}{a+b}\)

\(\Rightarrow S.\sqrt[3]{5}=\dfrac{\sqrt[3]{\left(a-2\right)\left(b-3\right).5}}{a+b}\)

\(\le\dfrac{\dfrac{\left(a-2\right)+\left(b-3\right)+5}{3}}{a+b}=\dfrac{\dfrac{a+b}{3}}{a+b}=\dfrac{1}{3}\)

\(\Rightarrow S\le\dfrac{1}{3}:\sqrt[3]{5}=\dfrac{1}{3\sqrt[3]{5}}\)

Đẳng thức xảy ra \(\Leftrightarrow a-2=b-3=5\Leftrightarrow\left\{{}\begin{matrix}a=7\\b=8\end{matrix}\right.\)