x+y+z=6 , xy+yz+zx=4 , 1/(1-x) + 1/(1-y) + 1/(1-z) =?
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anuveshkumars8338
24.01.2020
Math
Secondary School
+13 pts
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Xyz=1, prove that 1/(1+x+1/y)+1+(1+y+1/z)+1/(1+z+1/x)=1
2
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saurabhmehta9811
saurabhmehta9811 Virtuoso
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
Step-by-step explanation:
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