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Consecutive inequality (Posted on 2023-07-24)

Let x, y and z be consecutive integers such that

1/x+1/y+1/z>1/45

Find the maximum value of x+y+z.

No Solution Yet Submitted by Danish Ahmed Khan

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D1 solution

| Comment 2 of 3 |

If we call x=y-1 and z=y+1, the sum is going to be in the vicinity of 3/y but very slightly larger.

So y=45*3=135 and x+y+z=135*3=405.

In fact the difference between 1/x and 1/z is the tiny quantity 2/(y^2-1). If we go up to y=136 the sum is slightly larger than 3/136 = 1/(45+1/3).

Edited on July 25, 2023, 9:01 am

Posted by Jer on 2023-07-24 08:39:27

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