Let $p=(p_k)$ and $q=(q_k)$ be discrete probabilities. Let $q_{\max} > p_{\max}$. Then $H(q) < H(p)$?
I tried to prove it algebraically but failed. Is it supposed to be difficult?
Note for clarification that $p=(p_k)_{k=1}^K$, $\sum_k p_k = 1$ with $1 \ge p_k\geq 0$, and $H(p) = -\sum_k p_k \log p_k$.