X,y,z,m real numbers are so positive in which. Xyz=1 & 2xm/xy+x+1=2ym/yz+y+1=2zm/xz+z+1=1. Prove that m=1/2 ?

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Answer 1 · 2018-07-29T18:53:55

Please see Explanation.

#xyz=1 :. xy=1/z#.

Utilising this in #(2xm)/(xy+x+1)=1,# we get,

#(2xm)/(1/z+x+1)=1, or, (2xzm)/(1+xz+z)=1#.

Since, #(2zm)/(xz+z+1)=1...........[Given]#, we have,

# (2xzm)/(1+xz+z)=1=(2zm)/(xz+z+1), i.e., #

# x=1#.

Similarly, we can prove #y=1 and z=1#.

Thus, altogether, #x=y=z=1#.

Subst.ing these in #(2xm)/(xy+x+1)=1#, we find,

#m=(xy+x+1)/(2x)=(1+1+1)/2=3/2#.