Answer to Question #125496 in Algebra for Preeti Sarkar

Question

Find all integer x, y and z that satisfies both inequalities x>y>z >1 and (1/x) +(1/y) +(1/z) >1

Answer 1

1/x + 1/y + 1/z >1

the LCM is xyz

Therefore.. (yz + xz + xy)/xyz = 1

xy + yz + zx = xyz...............(1)

Now let, x = y = z = k (constant)

The value of x, y and z now on eqution (1) will be.

3K² = K³

3 = K

Hence x = y = z = 3

Then (x, y, z) = (3, 3, 3) (but this does not satisfy x > y > z > 1 hence it is not the answer ).

But since x > y > z > 1

If z = 2 and y = 3 (since z must be greater than one and y be greater than z) , then from equation (1)

(x)(3) + (3)(2) + (2)(x) =(x)(3)(2)

3x + 6 + 2x = 6x

5x + 6 = 6x

6 = x

Hence x = 6, when y = 3 and z = 2

(x, y, z) = (6, 3, 2) for it satisfies both x>y>z>1 and 1/x + 1/y + 1/z > 1

And If z = 2 and y = 4 , then from equation (1)

(x)(4) + (4)(2) + (2)(x) =(x)(4)(2)

4x + 8 + 2x = 8x

6x + 8 = 8x

8 = 2x

4 = x

hence (x, y, z) = (4, 4, 2) not the answer for the values does not satisfy (x > y > z > 1)

Therefore the only correct answer is (x, y, z )=( 6, 3 , 2)