$\def\N{\mathbb{N}}\def\R{\mathbb{R}}\def\paren#1{\left(#1\right)}$Lemma 1: Suppose $a_1, \cdots, a_n, u_1, \cdots, u_n \in \R$ and $\sum\limits_{k = 1}^n a_k = 1$. If$$
\sum_{k = 1}^n a_k t^{u_k} \geqslant 1,\quad \forall 0 < t \leqslant 1
$$
then $\sum\limits_{k = 1}^n a_k u_k \leqslant 0$.
Proof: Define $F(t) = \sum\limits_{k = 1}^n a_k t^{u_k}$. Note that $F'(1)$ exists and $F(1) = \sum\limits_{k = 1}^n a_k = 1$. Suppose $F'(1) > 0$, there exists $0 < δ < 1$ such that $F(t) < F(1) = 1$ for $t \in (1 - δ, 1)$, a contradiction. Thus $F'(1) \leqslant 0$, i.e.\begin{gather*}
F'(1) = \left.\paren{ \sum_{k = 1}^n a_k u_k t^{u_k - 1} }\right|_{t = 1} = \sum_{k = 1}^n a_k u_k \leqslant 0. \tag*{$\Box$}
\end{gather*}
Now return to the question. The proposition to be proved can be formalized as below (@RiverLi):
Proposition: Suppose $m, n \in \N_+$. For any $a_1, \cdots, a_n, b_1, \cdots, b_n, c_1, \cdots, c_n \in \R$, if $\sum\limits_{k = 1}^n a_k = 1$, $b_k + c_k = -m$ for $1 \leqslant k \leqslant n$, and\begin{gather*}
\sum_{k = 1}^n a_k x^{b_k} y^{c_k} \geqslant \frac{1}{x^m},\quad \forall y \geqslant x \geqslant 1 \tag{1}
\end{gather*}
then $\sum\limits_{k = 1}^n a_k · (-b_k) \geqslant m$.
Proof: Since $b_k + c_k = -m$ for all $k$, then$$
x^m \sum_{k = 1}^n a_k x^{b_k} y^{c_k} = \sum_{k = 1}^n a_k x^{b_k + m} y^{c_k} = \sum_{k = 1}^n a_k \paren{ \frac{x}{y} }^{b_k + m},
$$
thus (1) is equivalent to$$
\sum_{k = 1}^n a_k t^{b_k + m} \geqslant 1.\quad \forall 0 < t \leqslant 1
$$
Therefore, lemma 1 implies that $\sum\limits_{k = 1}^n a_k (b_k + m) \leqslant 0$, then\begin{gather*}
-\sum_{k = 1}^n a_k b_k \geqslant m \sum_{k = 1}^n a_k = m. \tag*{$\Box$}
\end{gather*}
In fact, the condition in lemma 1 is also necessary when $a_1, \cdots, a_n > 0$.
Lemma 2: Suppose $a_1, \cdots, a_n > 0$, $u_1, \cdots, u_n \in \R$. If $\sum\limits_{k = 1}^n a_k = 1$ and $\sum\limits_{k = 1}^n a_k u_k \leqslant 0$, then$$
\sum_{k = 1}^n a_k t^{u_k} \geqslant 1.\quad \forall 0 < t \leqslant 1
$$
Proof: Because $\sum\limits_{k = 1}^n a_k = 1$, the AM-GM inequality implies that for $0 < t \leqslant 1$,\begin{gather*}
\sum_{k = 1}^n a_k t^{u_k} \geqslant \prod_{k = 1}^n (t^{u_k})^{a_k} = t^{\sum\limits_{k = 1}^n a_k u_k} \geqslant 1. \tag*{$\Box$}
\end{gather*}
