I understand that Fisher's LSD test is used as a post hoc analysis if the ANOVA test rejects the null hypothesis All the means $X_1, X_2,.. X_n$ taken from various samples are equal.
Also, given that if we reject null hypothesis via ANOVA, we can definitely find at least one pair of means $X_i \& X_j$ such that Fisher's LSD test is also violated. - This is what has been taught and proven in my stats course.
However, is it possible that we cannot reject the Null hypotheses via ANOVA and yet find of a pair $X_i , X_j$ such that Fisher's LSD will reject that both the samples have same mean. - I feel that this type of scenario is possible.
So, is Fisher's LSD ( for all the pairs) more stricter test compared to ANOVA for rejecting the null hypothesis?
edit: None of the post-hoc tests are guaranteed to find a significant pair even after rejecting main ANOVA. This is true with Fisher's test as described here, but also with a popular alternative like Turkey's test