The second last proof is not entirely complete as in each case an x was found by AP to satisfy the inquality, but it's not necessarily the same x, one has to take the max of the two x's found.
That crossed my mind too. But since y∈(3,7) by assumption, 1/(y-3) > 1/(7-y), so any x that is greater than or equal to 1/(y-3) should do the trick imo.
The second last proof is not entirely complete as in each case an x was found by AP to satisfy the inquality, but it's not necessarily the same x, one has to take the max of the two x's found.
That crossed my mind too. But since y∈(3,7) by assumption, 1/(y-3) > 1/(7-y), so any x that is greater than or equal to 1/(y-3) should do the trick imo.
@@HC83KIm y∈(3,7) does not imply that 1/(y-3) > 1/(7-y)