Question

cho \(x\ge1,y\ge2,z\ge3\)

tìm GTLN của  \(A=\dfrac{yz\sqrt{x-1}+xz\sqrt{y-2}+xy\sqrt{z-3}}{xyz}\)

Answer 1

\(=>A=\dfrac{\sqrt{x-1}}{x}+\dfrac{\sqrt{y-2}}{y}+\dfrac{\sqrt{z-3}}{z}\)

áp dụng BĐT AM-GM

\(=>\sqrt{x-1}\le\dfrac{x-1+1}{2}=\dfrac{x}{2}\)

\(=>\dfrac{\sqrt{x-1}}{x}\le\dfrac{\dfrac{x}{2}}{x}=\dfrac{1}{2}\left(1\right)\)

có \(\dfrac{\sqrt{y-2}}{y}=\dfrac{\sqrt{\left(y-2\right)2}}{\sqrt{2}.y}\)

\(=>\sqrt{\left(y-2\right)2}\le\dfrac{y-2+2}{2}=\dfrac{y}{2}\)

\(=>\dfrac{\sqrt{\left(y-2\right)2}}{\sqrt{2}.y}\le\dfrac{\dfrac{y}{2}}{\sqrt{2}.y}=\dfrac{1}{2\sqrt{2}}\left(2\right)\)

tương tự \(=>\dfrac{\sqrt{z-3}}{z}\le\dfrac{1}{2\sqrt{3}}\left(3\right)\)

(1)(2)(3)\(=>A\le\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}\)