N Variable Constrained Minimization Problem Given $ \sum_{i=1}^{n} b_i \log(b_i) = K$ and $ \sum_{i=1}^{n} b_i = 1$ , what is $ \underset{i}{\min} (b_i) $? 
Note : $b_i \geq 0 \ \forall i$. Here $K$ is a constant. 
The value of $K$ obviously plays a big role ( for example $K$ has to be greater than $-\log(n)$. I am trying to minimize(or bound) the value of $b_i$. 
I have looked into conditions involving KKT and Lagrange multipliers, but none of them talk about minimizing across n variables. I don't even know how to approach the problem.
 A: If $n\leq2$ there is nothing to minimize. Therefore assume $n\geq3$. It is sufficient to determine the minimum possible value of $b_1$. We put
$$b_1:=x\leq{1\over n}, \qquad b_i:=(1-x)p_i\quad(2\leq i\leq n)$$
with $p_i\geq0$, $\ \sum_{i=2}^n p_i=1$. The main condition then is
$$\sum_{i=1}^nb_i\log{1\over b_i}=x\log{1\over x}+(1-x)\sum_{i=2}^n p_i\left(\log{1\over p_i}+\log{1\over 1-x}\right)=K\ ,$$
or
$$f(x,y):=x\log{1\over x}+(1-x)\log{1\over 1-x}+(1-x) y-K=0\ ,\tag{1}$$
where we have put $$\sum_{i=2}^np_i\log{1\over p_i}=:y\ \in [0,\log(n-1)]\ .$$
If $0\leq K\leq\log(n-1)$ we can satisfy $(1)$ by choosing $y=K$ and $x=0$. The min in question then is $0$ in this case.
If $\log(n-1)<K\leq \log n$ we consider $(1)$ as an equation for $x$, given $y$. One computes
$$x'(y)=-{f_y(x,y)\over f_x(x,y)}=-{1-x\over h_x-y}\ ,$$
where
$$h(x):=x\log{1\over x}+(1-x)\log{1\over 1-x}\ .$$
An easy computation shows that $h'(x)\geq\log(n-1)$ when $x<{1\over n}$. It follows that $x'(y)<0$ as long as $y<\log(n-1)$. In order to minimize $x$ we therefore should choose $y=\log(n-1)$ in this case as well. The minimal $b_1$ then is the solution of the equation
$$x\log{1\over x}+(1-x)\log{n-1\over 1-x}=K\ .$$
