Question

if x y and z are real positive numbers such that xyz=1 then find the value of 1by 1+x+xy + 1 by 1+y+yz + 1 by 1+z+zx

Answer 1

Answer:

Let A=1+x+1yA=1+x+1y

Let B=1+y+1zB=1+y+1z

Let C=1+z+1xC=1+z+1x

You may want to multiply AA with yy to get:

Ay=y+xy+1Ay=y+xy+1

Since xyz=1xyz=1 , we know that xy=1zxy=1z

Therefore, it is easy to see that:

Ay=BAy=B

Using the similar reasoning, can you show that Bz=CBz=C and Cx=ACx=A ?

So:

1A+1B+1C=1A+1Ay+1Bz1A+1B+1C=1A+1Ay+1Bz

=1A+1Ay+1Ayz=1A+1Ay+1Ayz

=yz+z+1Ayz=yz+z+1Ayz

=1x+z+1Bz=1x+z+1Bz

=CC=1=CC=1

Answer 2

Given :- x y and z are real positive numbers such that xyz = 1 .

To Find :- 1/(1 + x + xy) + 1/(1 + y + yz) + 1/(1 + z + zx) = ?

Solution :-

→ 1/(1 + x + xy) + 1/(1 + y + yz) + 1/(1 + z + zx)

Multiply and divide second term by x and third term by xy we get,

→ 1/(1 + x + xy) + x/x(1 + y + yz) + xy/xy(1 + z + zx)

→ 1/(1 + x + xy) + x/(x + xy + xyz) + xy/(xy + xyz + x²yz)

putting :-

• xyz = 1 in second term denominator .
• xyz = 1 in third term denominator
• x²yz = x(xyz) = x in third term denominator

→ 1/(1 + x + xy) + x/(1 + x + xy) + xy/(1 + x + xy)

taking LCM now,

→ (1 + x + xy) / (1 + x + xy)

→ 1 (Ans.)

Hence, required result is equal to 1 .

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