x,y,z,1／x＋1／y＋1／z＝2,求证:√(x+y+z)≥√(x－1)+√(y－1)+√(z－1)

Question

x,y,z,1／x＋1／y＋1／z＝2,求证:√(x+y+z) ≥√(x－1)+ √(y－1)+ √(z－1)

Answer 1

由1/x+1/y+1/z=2
得(x-1)/x+(y-1)/y+(z-1)/z=1
所以（x+y+z）=（x+y+z）[(x-1)/x+(y-1)/y+(z-1)/z]
=x-1 + y-1 + z-1 +(x-1)(y+z)/x+(y-1)(x+z)/y+(z-1)(x+y)/z
=x-1 + y-1 + z-1 + (x-1)y/x +(y-1)x/y +(y-1)z/y+(z-1)y/z +(z-1)x/z+(x-1)z/x
≥x-1 + y-1 + z-1 + 2√(x-1)(y-1) +2√(y-1)(z-1) +2√(z-1)(x-1)
=[√(x-1)+√(y-1)+√(z-1)]^2
即：√（x+y+z）≥√(x-1)+√(y-1)+√(z-1)