Focus on the endpoints of your current range [i,j]. If s[i]=s[j]: both characters can be endpoints of a palindrome. The remaining problem is finding the LPS in s[i+1..j−1], then adding 2 for these matching endpoints.
If s[i]=s[j]: at least one endpoint must be excluded. Either skip s[i] (solve s[i+1..j]) or skip s[j] (solve s[i..j−1]). Take the maximum. This gives us a clean two-case recurrence with no need to try all split points.