On the third edition of Ahlfors' Complex Analysis, page 110 Theorem 2(Cauchy's Theorem for a Rectangle) it states: It follows from (13) that at least one of the rectangles $R^{(k)},k=1,2,3,4$, must satisfy the condition $|\eta(R^{(k)})|\geq \frac{1}{4}|\eta(R)|$.
I'm confused about it.
We know that $\eta(R)=\int_{\partial R}f(z)dz$ and $\eta(R^{(k)})=\int_{\partial R^{(k)}}f(z)dz$.
It is also easy to know that $\eta(R)=\eta(R^{(1)})+\eta(R^{(2)})+\eta(R^{(3)})+\eta(R^{(4)})$. Let $\eta(R^{(k)})=a_k+ib_k$ and $|\eta(R^{(k)})|=\sqrt{a_k^2+b_k^2}$. Then we have $\eta(R)=\left(\sum_{k=1}^{4}a_k\right)+i\left(\sum_{k=1}^{4}b_k\right)$ and $|\eta(R)|=\sqrt{\left(\sum_{k=1}^{4}a_k\right)^2+\left(\sum_{k=1}^{4}b_k\right)^2}$.
How to prove at least one of the rectangles $R^{(k)},k=1,2,3,4$, must satisfy the condition $|\eta(R^{(k)})|\geq \frac{1}{4}|\eta(R)|$?
This a question from Ahlfors' Complex Analysis. I show you the whole context of the question below. I have trouble understanding the statement with red line.
The statement follows from integral additivity $(\star)$ and triangle inequality $(\star\star)$:
$$|\eta(R)|=\Bigl|\int_{\partial R}f\Bigr|\overset{(\star)}{=}\Bigl|\sum_{k=1}^4\int_{\partial R^{(k)}}f\Bigr|\overset{(\star\star)}{\leq}\sum_{k=1}^4\Bigl|\int_{\partial R^{(k)}}f\Bigr|=\sum_{k=1}^4 |\eta(R^{(k)})|.$$
Now let $h$ be such that $|\eta(R^{(h)})|=\max_k|\eta(R^{(k)})|$. In other words, pick the greatest of the four elements $|\eta(R^{(1)})|$, $|\eta(R^{(2)})|$, $|\eta(R^{(3)})|$, $|\eta(R^{(4)})|$.
Then
$$\sum_{k=1}^4 |\eta(R^{(k)})|\leq 4|\eta(R^{(h)})|.$$
Putting all together, we have
$$|\eta(R)|\leq 4|\eta(R^{(h)})|.$$