A problem about gradient estimates for Poisson's equation

Question

During studying Gilbarg-Trudinger's Elliptic PDEs of $2^{nd}$ order, I've been confused about an derivation of the estimates inequality for a long time. In sec. 3.4 Gradient Estimates for Poisson's Equation, they derived the estimate (3.16) by the following inequalities:
$d_{x}|Du(x)|\leq C(\sup_{\partial Q}|u|+d^2 \sup_{Q}|f|)\leq C(\sup_{\Omega}|u|+\sup_{\Omega}\ d^2_y|f(y)|) \tag{1}$
where $\Omega$ is any region (open and connected) in $\Bbb{R}^{n},\ d_{x}=\operatorname{dist}(x,\partial \Omega)$ and $d=d_{x}/\sqrt{n}$.
The former parts in the second and third formula is clear. Since $Q\subset\Omega$ and $u$ is continuous on $\bar{\Omega}$, $\sup_{\partial Q}|u|=\sup_{Q}|u|\leq \sup_{\Omega}|u|$.
But it seems no evidence that the formula holds in the latter parts, i.e.
$Cd^2 \sup_{Q}|f|\leq C\sup_{\Omega}\ d^2_y|f(y)| \tag{2}$
Since $d^2 \sup_{Q}|f|=\frac{d_x^2}{n} \sup_{Q}|f|$ and $\frac{1}{n}$ can be absorbed into the second $C$ in $(2)$. Therefore, the real matter is whether the following inequality holds:
$d_x^2 \sup_{Q}|f|\leq \sup_{\Omega}\ d^2_y|f(y)| \tag{3}$
For a simple counter example, by taking $x\in \Bbb{R}^n,\ \Omega=\{x+w: |w_i|<d_0,\ w=(w_1,...,w_n)\}$, we then get $Q=\Omega,\ d=d_0$. Since for any $y\in\Omega,\ d_x=d_0\geq d_y$, we have $d_x^2 \sup_{Q}|f|=d_0^2\sup_{\Omega}|f|\geq d_0^2|f(y)|\geq d_y^2|f(y)|$.
Therefore, $d_x^2 \sup_{Q}|f|\geq \sup_{\Omega}d_y^2|f(y)|$. And I get an inversed ordering of $(3)$.
I know I can't say $(1)$ is wrong if $(3)$ is wrong. But $(3)$ is the way I try to understand the proof of $(1)$. Can anyone explain why $(1)$ is correct which is written in the Gilbarg-Trudinger's book. Many thanks.