Expectation of Reciprocal of Sum of i.i.ds

Question

$X_i$s are real continous i.i.ds. I am trying to prove the conjecture

$$ E[\frac{1}{\sum_{i=1}^nX_i}] = E[\frac{1}{nX_i}] $$

I can't seem to be getting anywhere. Prove or disprove.

Answer 1

Arithmetic mean is greater than or equal to the harmonic mean.

$${1 \over n}\sum_{i=1}^n X_i \geq {n \over \sum_{i=1}^n {1 \over X_i}}$$ $$\sum_{i=1}^n {1 \over X_i} \geq {n^2 \over \sum_{i=1}^n X_i}$$ $$n \mathbb{E} \left[ {1 \over X_1} \right] \geq n^2 \mathbb{E}\left[ {1 \over \sum_{i=1}^n X_i} \right] $$ $$\mathbb{E}\left[ {1 \over \sum_{i=1}^n X_i} \right] \leq {1 \over n}\mathbb{E} \left[ {1 \over X_1} \right]$$