Cho x + y + z = 1
C/m: x/(x+1) + y/(y)+1 + z/(z+1) \leq 3/4
Đặt A= [TEX]\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}[/TEX]
Có: [TEX]A+ \frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=3.[/TEX]
Mà ta có BĐT [TEX](a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \geq 9[/TEX](cái này cm dùng cô si 3 số)
[TEX]\Rightarrow (\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1})(x+1+y+1+z+1) \geq 9[/TEX]
[TEX]\Leftrightarrow (\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1})(1+3) \geq 9[/TEX]
[TEX]\Leftrightarrow (\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}) \geq \frac{9}{4}[/TEX]
[TEX]\Rightarrow 3 = A+ \frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1} \geq A+ \frac{9}{4} \Leftrightarrow A \leq \frac{3}{4}[/TEX](đpcm)
Đẳng thức xảy ra khi [tex]a=b=c=\frac{1}{3}[/tex]