Question

Tìm GTLN:A=\(\dfrac{yz\sqrt{x-1}+xz\sqrt{y-4}+xy\sqrt{z-9}}{xyz}\)

Answer 1

đk: x≥1; y≥4; z≥9

Ta có: \(A=\dfrac{yz\sqrt{x-1}+xz\sqrt{y-4}+xy\sqrt{z-9}}{xyz}\)

\(=\dfrac{\sqrt{x-1}}{x}+\dfrac{\sqrt{y-4}}{y}+\dfrac{\sqrt{z-9}}{z}\)

A/dụng bđt côsi cho 2 số không âm \(\sqrt{x-1}\) và 1 có:

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

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

Tương tự: \(\dfrac{\sqrt{y-4}}{y}\le\dfrac{1}{4};\dfrac{\sqrt{z-9}}{z}\le\dfrac{1}{6}\)

Cộng theo vế các Bđt trên ta có:

\(\dfrac{\sqrt{x-1}}{x}+\dfrac{\sqrt{y-4}}{y}+\dfrac{\sqrt{z-9}}{z}\le\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}\)

hay \(A\le\dfrac{11}{12}\)

Dấu ''='' xảy ra khi x = 2; y = 8; z = 18

vậy max_A = \(\dfrac{11}{12}\)