Question

cho A=\(\dfrac{\sqrt{x-3}}{x}+\dfrac{\sqrt{y-4}}{y}+\dfrac{\sqrt{z-5}}{z}\) tìm giá trị lớn nhất

Answer 1

Answer 2

Lời giải:

Áp dụng BĐT Cô-si:

\(\sqrt{x-3}=\frac{1}{\sqrt{3}}.\sqrt{3(x-3)}\leq \frac{1}{\sqrt{3}}.\frac{3+(x-3)}{2}=\frac{x}{2\sqrt{3}}\)

\(\Rightarrow \frac{\sqrt{x-3}}{x}\leq \frac{1}{2\sqrt{3}}\)

\(\sqrt{y-4}=\frac{1}{2}\sqrt{4(y-4)}\leq \frac{1}{2}.\frac{4+(y-4)}{2}=\frac{y}{4}\)

\(\Rightarrow \frac{\sqrt{y-4}}{y}\leq \frac{1}{4}\)

\(\sqrt{z-5}=\frac{1}{\sqrt{5}}\sqrt{5(z-5)}\leq \frac{1}{\sqrt{5}}.\frac{5+(z-5)}{2}=\frac{z}{2\sqrt{5}}\)

\(\Rightarrow \frac{\sqrt{z-5}}{z}\leq \frac{1}{2\sqrt{5}}\)

Vậy \(A\leq \frac{1}{2\sqrt{3}}+\frac{1}{4}+\frac{1}{2\sqrt{5}}=A_{\max}\)

Dấu "=" xảy ra tại \(x=6; y=8; z=10\)