Given Integer $N,M$ with $1\leq M\leq N$, consider the following optimization problem w.r.t $P = (p_1,p_2,\cdots,p_{N+M})$
Constraints： $$\sum_{k=1}^{N+M} -p_k\log_2p_k-(1-p_k)\log_2(1-p_k) \geq N,\\\forall 1\leq k\leq N+M,\ 0\leq p_k\leq 0.5\\ $$ $$Minimize：\sum_{k=1}^{N+M} p_k(1-p_k)$$ We define $0\log_2 0 = 0$.
True or false:
The optimal $P^*$ that minimize the objective function will have all coordinates equal. i.e.
$P^* = (p,p,p,\cdots,p)$
My thoughts:
By intuition, the conjecture is true by symmetry.
I found both $F(p)=-p\log_2 p -(1-p)\log_2(1-p)$ and $G(p)=p(1-p)$ are monotonic and concave in range$[0,0.5]$. But I can not make any more progress from it.
I examined the case N=1, M=1; and N=2, M=1, and found the conjecture is true for both cases.
For a convex problem, one way to demonstrate the optimality of a solution is via the KKT conditions. Such a problem would be \begin{eqnarray} \text{minimise}&\quad&\sum_{k}g\left(y_{k}\right)\\ \text{subject to}&\quad&N-\sum_{k}f\left(y_{k}\right)\le 0 \end{eqnarray} with $g$ convex and $f$ concave.
By your definition $G$ is not a convex function. But we may construct $g$ convex (actually affine) by setting $g\left(z\right)=G\left(p\left(z\right)\right)$ with $p\left(z\right)=\left(1-\sqrt z\right)/2$. Then $$ g\left(z\right)=\left(1-z\right)/4. $$ The Lagrangian $$ L\left(y,\mu\right)=\sum_{k}g\left(y_{k}\right)+\left(N-\sum_{k}f\left(y_{k}\right)\right)\mu $$ has derivative with respect to $y_{k}$ $$ L_{y_{k}}\left(y,\mu\right)=-\frac{1}{4}-\mu\,f'\left(y_{k}\right). $$ To claim optimality it will suffice to exhibit $\left(y,\mu\right)$ such that $$ N-\sum_{k}f\left(y_{k}\right)=0 \quad\text{and}\quad \mu\ge 0 \quad\text{and}\quad -\frac{1}{4}-\mu\,f'\left(y_{k}\right)=0 \quad\text{for all }k. $$ The first condition implies primal feasibility, the second dual feasibility, the first (again) complementary slackness and the third that the gradient of $L$ vanishes. The KKT conditions will have been satisfied.
Such a $\left(y,\mu\right)$ may be constructed with $y=\left(z,z,\ldots,z\right)$ and $\mu=-1/\left(4f'\left(z\right)\right)$ where we have $z$ satisfying $$ f\left(z\right)=\frac{N}{N+M} \quad\text{and}\quad f'\left(z\right)<0. $$ Of course we define $f(z)=F\left(p\left(z\right)\right)$. So $$ f(z)= 1-\frac{\left(1-\sqrt z\right) \ln\left(1-\sqrt z\right) +\left(1+\sqrt z\right) \ln\left(1+\sqrt z\right)} {2\ln 2} $$ $$ f'(z)= -\frac{\ln\left(1+\sqrt z\right)-\ln\left(1-\sqrt z\right)}{\left(4\ln 2\right)\sqrt{z}} = -\frac{\text{arctanh}\sqrt z}{\left(2\ln 2\right)\sqrt{z}} $$ $$ f''\left(z\right)= \frac{\left(1-z\right)\text{arctanh} \sqrt z -\sqrt z} {\left(4\ln 2\right)\left(1-z\right)z^{3/2}}. $$ With domain $\left(0,1\right)$, $f$ has range $\left(0,1\right)$. So $0<N/\left(N+M\right)<1$ implies some $z$ with $0<z<1$ satisfying $f\left(z\right)=N/\left(N+M\right)$. And clearly $f'\left(z\right)<0$.
It remains to show that $f$ is indeed concave, using its second derivative. Almost certainly there are neater ways to do it but here is mine. For $x > 0$, $$ 1 < \cosh x = \frac{d}{dx}\left(\sinh x\right) $$ $$ x \text{ sech} x < x < \sinh x $$ $$ \left(1-\tanh^{2}x\right) x = x \text{ sech}^{2} x < \tanh x $$ Since $0<z<1$ implies $z=\tanh^{2}x$ for some $x>0$, $$ \left(1-z\right)\text{arctanh} \sqrt z < \sqrt z $$ $$ \frac{\left(1-z\right)\text{arctanh} \sqrt z -\sqrt z} {\left(4\ln 2\right)\left(1-z\right)z^{3/2}}< 0. $$
The conjecture is true, but we needed the convexity to prove it. An appeal to symmetry is not enough.
Here are some plots about $f$. I have more confidence in these than I have in my algebraic manipulations!