Probability on one member of a partition of the sample space

This question is exercise 18, chapter 1 of All of Statistics

Suppose $k$ events form a partition of the sample $\Omega$, i.e. they are disjoint and $\cup_{i = 1}^{k} A_i = \Omega$. Assume that $P(B) > 0$. Prove that if $P(A_1|B) < P(A_1)$ then $P(A_i|B) > P(A_i)$ for some $i = 2,..., k$

My current approach, which might not be correct, is trying to prove that $\Pi_{i=1}^{k}\frac{P(A_i|B)}{P(A_i)} = 1$, and since $\frac{P(A_1|B)}{P(A_1)} < 1$ there should be at least one $\frac{P(A_i|B)}{P(A_i)} > 1$ for $i = 2,..., k$

Suppose that $$\mathbb{P}(A_1|B)<\mathbb{P}(A_1)\Longleftrightarrow\frac{\mathbb{P}(A_1\cap B)}{\mathbb P(B)}<\mathbb P(A_1)$$ but also that $$\mathbb{P}(A_j|B)\leq\mathbb{P}(A_j)\Longleftrightarrow\frac{\mathbb{P}(A_j\cap B)}{\mathbb P(B)}\leq\mathbb P(A_j)$$ for all $j\in\{2,\ldots,k\}$. Since the $A_j$'s are disjoint and their union is the whole space, we have that \begin{align*}\mathbb{P}(B)=&\,\sum_{j=1}^k\mathbb P(A_j\cap B)=\mathbb P(A_1\cap B)+\sum_{j=2}^k\mathbb P(A_j\cap B)<\mathbb{P}(A_1)\mathbb P(B)+\sum_{j=2}^k \mathbb P(A_j)\mathbb P(B)\\ =&\,\left(\sum_{j=1}^k \mathbb P(A_j)\right)\,\mathbb P(B)=\mathbb P(B),\end{align*} which is a contradiction. Hence, if $\mathbb{P}(A_1|B)<\mathbb{P}(A_1)$, then $\mathbb{P}(A_j|B)>\mathbb{P}(A_j)$ for at least one $j\in\{2,\ldots,k\}$.