Question

Cho x,y,z > 0 và x+y+z=1.Tìm giá trị nhỏ nhất của \(X=\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\left(1+\dfrac{1}{z}\right)\)

Answer 1

Chừ ms onl nên ko bt

Ta có: \(X=\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\left(1+\dfrac{1}{z}\right)\)

\(=1+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\left(\dfrac{1}{yz}+\dfrac{1}{xz}+\dfrac{1}{xy}+\dfrac{1}{xyz}\right)\)

\(\ge1+\dfrac{9}{x+y+z}+\left(\dfrac{x+y+z}{xyz}+\dfrac{1}{xyz}\right)\)

\(=10+\dfrac{2}{xyz}\) ( Do \(x+y+z=1\) )

Áp dụng BĐT AM-GM ta có:

\(\left(\dfrac{x+y+z}{3}\right)^3\ge xyz\) \(\Leftrightarrow\dfrac{1}{xyz}\ge27\)

\(\Rightarrow X\ge10+27.2=64\)

\(\Rightarrow\) Dấu ''='' xảy ra \(\Leftrightarrow x=y=z=\dfrac{1}{3}\)