What is the tightest upper bound one can obtain for the following expression


$$\sum\_{i=1}^kA\_i\log(\frac{A\_i}{e})$$ subject to $\sum\_{i = 1}^k A\_i = C$ in terms of $C$ and $k$?


A very loose upper bound for this expression is $C\log(\frac{C}{e})$. Can we do better than this?



Tighter upper bound for $\sum_{i=1}^kA_i\log(\frac{A_i}{e})$
The gradient of your function is $(\log A\_1, \dotsc, \log A\_n).$ The function is convex, and so the minimum can be seen, by Lagrange multipliers to occur when all of the $A\_i$ are equal (to $C/k$). So, the minimum value of the function is 
$$C \log\left(\frac{C}{ke}\right).$$


On the other hand, the maximum must occur on the boundary, which is to say, when some of the $A\_i$ are $0.$ It is clear that in fact for a maximum, *all but one* $A\_i$ are $0,$ on which case (since $\lim\_{x\to 0} x \log x = 0$), the value is $C \log\frac{C}{e}.$.


