Question

x+y+z=1,x/(1+x)+y/(1+y)+z/(1+z)

Answer 1

x/(1+x)+y/(1+y)+z/(1+z)≤3/4

即1-1/(1+x)+1-1/(1+y)+1-1/(1+z)≤3/4

即1/(1+x)+1/(1+y)+1/(1+z)≥9/4

即4[1/(1+x)+1/(1+y)+1/(1+z)]≥9

即[(1+x)+(1+y)+(1+z)][1/(1+x)+1/(1+y)+1/(1+z)]≥9

由于(1+x)+(1+y)+(1+z)≥3倍的3次根号下[(1+x)(1+y)(1+z)]

1/(1+x)+1/(1+y)+1/(1+z)≥3倍的3次根号下[1/(1+x)(1+y)(1+z)]

所以[(1+x)+(1+y)+(1+z)][1/(1+x)+1/(1+y)+1/(1+z)]≥9

所以原不等式成立.