Question

if X,y,z are real positive ,then prove that : √xy^-1 . √yz^-1 . √zx^-1 = 1
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Answer 1

Answer:

Given xyz = 1

xy = 1/z, yz = 1/x , xz = 1/y

Expression to be evaluated is 1/ (1 + x + 1/y) + 1 / (1 + y + 1/z) + 1/ (1 + z + 1/x)

Now consider

1/ (1 + x + 1/y) = y / (y + xy + 1) ------------- (1

= y / (y + 1/z + 1) = yz / (yz + 1 + z)

= yz / ( 1/x + 1 + z) = (1/x)/ ( 1/x + 1 + z)

From above we get (using ratio and proportion a/c= b/d = (a+b)/(c+d)]

1/ (1 + x + 1/y) = y / (y + 1/z + 1) = (1/x) / (1/x + 1 + z) = [1+ y + 1/x] / [(1 + x + 1/y) + (y + 1/z + 1) + ( 1/x + 1 + z)] --------------------------1

Or

1/ (1 + x + 1/y) = [1+ y + 1/x] / [(1 + x + 1/y) + (y + 1/z + 1) + ( 1/x + 1 + z)] -------(A)

Replacing x by y and y by z and z by x we get

1/ (1 + y + 1/z) = [1+ z + 1/y] / [(1 + y + 1/z) + (z + 1/x + 1) + (1/y + 1 + x)] -------(B)

And again x by y and y by z and z by x we get

1/ (1 + z + 1/x) = [1+ x + 1/z] / [(1 + z + 1/x) + (x + 1/y + 1) + (1/z + 1 + y)] --------- ( C)

We observe that in all cases A, B , C the denominator is same or = [(1+1+1) + (x + y + z) + (1/x + 1/y +1/z)]

Now adding LHS of (A + B + C) = Adding RHS OF (A + B + C)

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

= {[1+ y + 1/x] +[1+ z + 1/y] + [1+ x + 1/z]} / [(1+1+1) + (x + y + z) + (1/x + 1/y +1/z)]

= 1

Hence the given expression has value = 1