Question

cho x,y,x>0

cm: \(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}< =\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)

Answer 1

Lời giải:Áp dụng BĐT Cauchy-Schwarz ta có:

$\frac{1}{2x+y+z}\leq \frac{1}{16}\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$

$\frac{1}{x+2y+z}\leq \frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}\right)$

$\frac{1}{x+y+2z}\leq \frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{z}\right)$

Cộng theo vế và rút gọn thì:

$\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\leq \frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$