Let $a_1,\ldots,a_n, A$ be positive real numbers and $k$ a positive integer. Is this correct? If we have $$\sum_{i=1}^n\ln\left(1+a_i\right)\geq k\ln\left(1+A/k\right),$$ and $$\sum_{i=1}^na_i\leq A,$$ then $$n=k,$$ and $$a_i=A/k.$$
If it is not correct, how can I choose the right hand side of the first inequality to make it correct?
Try some examples. Let's say $k=1$. The first inequality says $\sum_{i=1}^n \ln(1+a_i) \ge \ln(1+A)$, or $\prod_{i=1}^n (1+a_i) \ge 1 + A$, the second says $\sum_{i=1}^n a_i \le A$. As long as $\prod_{i=1}^n (1+a_i) \ge 1 + \sum_{i=1}^n a_i$, there will be some $A$ that makes this true. But in fact that's always true if all $a_i > 0$, because when you expand out the product the terms on the right side will all appear, together with some other positive terms.
Again, let's say $a_i = t b_i$ and $A = \sum_{i=1}^n a_i = t \sum_{i=1}^n b_i $. Exponentiating the first inequality gives us $$\prod_{i=1}^n (1 + t b_i) \ge \left(1 + (t/k) \sum_{i=1}^n b_i\right)^k$$ As polynomials in $t$, the left side has degree $n$ and the right side has degree $k$. If $n > k$, for any positive $b_i$ this will always be true if $t$ is large enough.