Question

cho a ≥ 3, b ≥ 4,c ≥ 2 tìm max P=\(\dfrac{ab\sqrt{c-2}+bc\sqrt{a-3}+ca\sqrt{b-4}}{abc}\)

Answer 1

\(\Leftrightarrow P=\dfrac{\sqrt{c-2}}{c}+\dfrac{\sqrt{a-3}}{a}+\dfrac{\sqrt{b-4}}{b}\)

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

\(dấu"="xảy\) \(ra\Leftrightarrow\left\{{}\begin{matrix}3=a-3\\4=b-4\\2=c-2\\\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=6\\b=8\\c=4\end{matrix}\right.\)