Question

Cho x,y,z>1 và \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=2\)

Chứng minh \(\sqrt{x+y+z}\ge\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}\)

Answer 1

Xét \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=2\)

\(\Leftrightarrow1=\left(1-\dfrac{1}{x}\right)+\left(1-\dfrac{1}{y}\right)+\left(1-\dfrac{1}{z}\right)\)

\(\Leftrightarrow1=\dfrac{x-1}{x}+\dfrac{y-1}{y}+\dfrac{z-1}{z}\)

Áp dụng bđt Bunhiacopxki có:

\(x+y+z=\left(x+y+z\right)\left(\dfrac{x-1}{x}+\dfrac{y-1}{y}+\dfrac{z-1}{1}\right)\ge\left(\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}\right)^2\)\(\Leftrightarrow\sqrt{x+y+z}\ge\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}\)

Dấu "=" xảy ra khi x=y=z=1,5Tự đăng câu hỏi xong tự trả lời  (T-T)