How to write the following channel capacity using CVX free error?

Convert log2(...) to log(...)/log(2).

I think the solution at Can CVX solve this kind of function {x-log(1+0.01*x/(x+1))} can be adapted to handle each of the log terms. You’ll learn more if you do the adapting than if I do it for you.

The min can be left as is.

The end result should be a concave expression.

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