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Stuart_Armstrong comments on Solving the two envelopes problem - Less Wrong
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Even simpler counter example: X and Y take values 1 and 100, independently and with equal probability. Then E(X/Y) = 1/4 ( 1/1 + 1/100 + 100/100 + 100/1), which is basically 25.5. Ditto for E(Y/X).
Expectation is a mean, and in means, large terms dominate - for E(X/Y), the (X=100,Y=1) situation dominates, while for E(Y/X), the (X=1,Y=100) situation dominates.
In fact, if X and Y are independent, identically distributed, strictly positive and non-trivial (ie not just constants), then I think that we always have E(X/Y) > 1.