Suppose $X_i$ are i.i.d. nonnegative random variables and $w_i$ and $v_i$ are nonnegative weights.
It is given that $\mathbb E \sum w_i X_i < \mathbb E \sum v_iX_i$. Then can we claim that $$\mathbb E \log(1+ \frac{1}{\sum v_iX_i})<\mathbb E \log(1+ \frac{1}{\sum w_iX_i}).$$
I tried with Jensen's inequality and could lover bound each side separately but couldn't establish the relationship I want.
I know that the expectations in $\mathbb E \sum w_i X_i < \mathbb E \sum v_iX_i$ cancels out as i.i.d. so $ \sum w_i < \sum v_i$. Which is equal to the condition in my previous question Expectation of a function of weighted sum of random variables: Is this coupling? But hoping this condition is stronger to work on.
Here is a rather laborious counterexample. There must be a simpler one...
Suppose $X_1,X_2 $ take values $1,2$ with probability ${1 \over 2}$. Let $w_1=w_2 = 1$ and $v_1 =t, v_2 = 2$, with $t >0$. Then we have $3=E [\sum_i w_i X_i] < E [ \sum_i v_i X_i ]=3 + { 3 t \over 2}$.
We have $E \log(1 + { 1 \over \sum_i w_i X_i}) = {1 \over 4} (\log {3 \over 2} + \log { 4 \over3} + \log { 4 \over3} + \log { 5 \over 4} )$ and $\epsilon_t = E \log(1 + { 1 \over \sum_i v_i X_i}) = {1 \over 4} (\log(1+ {1 \over t+2}) + \log(1+ {1 \over t+4}) + \log(1+ {1 \over 2t+2}) + \log(1+ {1 \over 2t+4}))$.
Note that $e_0 = {1 \over 4} (\log {3 \over 2} + \log { 5 \over 4} + \log { 3 \over 2} + \log { 5 \over 4} ) > E \log(1 + { 1 \over \sum_i w_i X_i})$, hence by continuity we have $E \log(1 + { 1 \over \sum_i w_i X_i}) < E \log(1 + { 1 \over \sum_i v_i X_i})$ for some small $t>0$.