Question

If xyz=1,then show that
(1+x+y-1)-1+(1+y+z-1)-1+(1+z+x-1)-1=1

Answer 1

Given xyz = 1

xy = 1/z, yz = 1/x , xz = 1/y 

Now {[1+x+y^-1]^-1}+{[1+y+z^-1]^-1}+{[1+z+x^-1]^-1}

⇒ [ 1+ x + 1/ y]-1 + [ 1 + y + 1 / z ] -1 + [ 1 + z + 1 /x ]-1

⇒ 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)

= 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)] --------------------------(A)

Or

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

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)] -------(2)

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)] --------- ( 3)

We observe that in all cases 1, 2 , 3 the denominator is same or = [(1+1+1) + (x + y + z) + (1/x + 1/y +1/z)]

Now adding LHS of (1 + 2 + 3) = Adding RHS OF (1 + 2 + 3)

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)]

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Answer 2

Answer:wtf

Step-by-step explanation:idk