ta có : \(\frac{x}{x+y}>\frac{x}{x+y+z}\left(1\right)\); \(\frac{y}{y+z}>\frac{y}{y+z+z}\left(2\right)\); \(\frac{z}{z+x}>\frac{z}{z+x+y}\left(3\right).\)
cộng vế với vế các BĐT (1), (2), (3) ta được:
\(\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}>\frac{x+y+z}{x+y+z}=1.\)(đpcm )
cái (2) gõ nhầm phím . nhé \(\frac{y}{y+z}>\frac{y}{y+z+x}\)
TA CÓ :\(\frac{X}{X+Y}\)>\(\frac{X}{X+Y+Z}\)(1) ; \(\frac{Y}{Y+Z}\)>\(\frac{Y}{X+Y+Z}\)(2) ; \(\frac{Z}{Z+X}\)>\(\frac{Z}{X+Y+Z}\)(3)
CỘNG (1) , (2) ,(3) vế theo vế ,ta có; \(\frac{X}{X+Y}\)+ \(\frac{Y}{Y+Z}\)+\(\frac{Z}{Z+X}\)> \(\frac{X+Y+Z}{X+Y+Z}\)= 1 (đpcm)
\(\frac{x}{x+y}>\frac{x}{x+y+z}\)(1)
\(\frac{y}{y+z}>\frac{y}{x+y+z}\)(2)
\(\frac{z}{z+x}>\frac{z}{x+y+z}\)(3)
Từ (1);(2);(3) suy ra ĐPCM