Question

Chứng minh:

+) (x+y+z)³ \(\ge27xyz\)

+) \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\)

Answer 1

\(\left(x+y+z\right)^3\ge\left(3\sqrt[3]{xyz}\right)^3=27xyz\) ( Cosi ) 

3 cách : 

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge3\sqrt[3]{\frac{1}{xyz}}=\frac{3}{\sqrt[3]{xyz}}\ge\frac{3}{\frac{x+y+z}{3}}=\frac{9}{x+y+z}\) ( Cosi 2 lần ) 

\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge3\sqrt[3]{xyz}.3\sqrt[3]{\frac{1}{xyz}}=9\)\(\Leftrightarrow\)\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\) ( Cosi 2 tích ) 

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{\left(1+1+1\right)^2}{x+y+z}=\frac{9}{x+y+z}\) ( Cauchy-Schwarz dạng Engel ) 

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