Question

Xyz=1, prove that 1/(1+x+1/y)+1+(1+y+1/z)+1/(1+z+1/x)=1

Answer 1

if xyz=1 then show that {[1+x+y^-1]^-1}+{[1+y+z^-1]^-1}+{[1+z+x^-1]^-1}=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)]

= 1

Answer 2

Step-by-step explanation:

Given: xyz is equals to 1

To Prove: 1/ ( 1+ x + 1/y ) + 1 / (1 + y + 1/z) + 1 (1 + z + 1/x ) /= 1

Proof:

》1/ (y + xy + 1/y ) + 1 /(1/z + yz + 1/z )+ 1 /(x + z + 1/x ) = 1

》y/(y +xy + 1) + z/(z + yz + 1) + x/(x + xz + 1) = 1

》1/ (1 + 1 + 1)+ 1/(1 +1+ 1)+1/(1 +1+ 1) = 1 <given ; xyz = 1 it can only be posssible when x=1,y=1 & z=1 >

》1/3 + 1/3 + 1/3 = 1

》3/3= 1

》1 = 1 < LHS = RHS >

Hence, proved.

plz mrk as brainliest if its helpful!.... thanks :)