Can you tell the difference?

Let the side length of the square be $1$ , the radii of the white, yellow and red circles be $r_1$ , $r_2$ , and $r_3$ respectively.

We note the side length of the square is given by:

$\begin{aligned} r_3 + \sqrt{(r_1+r_3)^2-(r_1-r_3)^2} + r_1 & = 1 \\ r_3 + 2 \sqrt{r_3r_1} + r_1 & = 1 \\ (\sqrt{r_3}+\sqrt{r_1})^2 & = 1 \\ \implies \sqrt{r_3}+\sqrt{r_1} & = 1 & ...(1) \end{aligned}$

The diagonal of the square is given by:

$\begin{aligned} (\sqrt 2+1)(r_2+r_1) & = \sqrt 2 \\ \implies r_2 & = 2 - \sqrt 2 - r_1 & ...(2) \end{aligned}$

The length of the black line segment is given by both $\ell = 2r_3 \cot \frac {135^\circ}2 + 4 \sqrt{r_3r_1}$ and $\ell = 2 r_2 \cot \frac {45^\circ}2$ . Therefore, we have:

$\begin{aligned} r_3 \cot \frac {135^\circ}2 + \blue{2 \sqrt{r_3r_1}} & = r_2 \cot \frac {45^\circ}2 & \small \blue{\text{Note that }r_3 + 2 \sqrt{r_3r_1} + r_1 = 1} \\ (\sqrt 2-1)r_3 + \blue{1-r_3-r_1} & = (\sqrt 2 + 1)r_2 \\ (\sqrt 2-2) r_3 + 1 - r_1 + \blue{(\sqrt 2+1)r_1} & = \blue{(\sqrt 2 +1)(r_2 + r_1)} & \small \blue{\text{and }(\sqrt 2+1)(r_2+r_1) = \sqrt 2} \\ (\sqrt 2-2) \blue{r_3} + 1 + \sqrt 2 r_1 & = \sqrt 2 & \small \blue{\text{and }\sqrt{r_3}+\sqrt{r_1} = 1 } \\ (\sqrt 2-2) \blue{(1-\sqrt{r_1})^2} + 1 + \sqrt 2 r_1 & = \sqrt 2 \\ (\sqrt 2-2) (1-2\sqrt{r_1}+r_1) + 1 + \sqrt 2 r_1 & = \sqrt 2 \\ 2(\sqrt 2 -1)r_1 + 2(2-\sqrt 2)\sqrt{r_1} - 1 & = 0 \\ \implies 2r_1 + 2\sqrt {2r_1} - \sqrt 2 - 1 & = 0 \\ \implies r_1 = \frac 32 + \frac 1{\sqrt 2} - \sqrt{2+\sqrt 2} & \approx 0.359347716164 \\ r_2 = \frac 12 - \frac 3{\sqrt 2} + \sqrt{2+\sqrt 2} & \approx 0.226438721463 & \small \blue{\text{From (2)}} \\ r_3 = \small \frac 52 + \frac 1{\sqrt 2} - \sqrt{2 + \sqrt 2} - \sqrt{2\left(3+\sqrt 2 - 2\sqrt{2+\sqrt 2}\right)} & \approx 0.160435348784 & \small \blue{\text{From (1)}} \end{aligned}$

Note that the area of the blue triangle is $\Delta_\blue{\rm blue} = (\sqrt 2+1)^2r_2^2$ and then the blue area $A_\blue{\rm blue} = \Delta_\blue{\rm blue} - \pi r_2^2 \approx 0.137766079523$ and the green area $A_\green{\rm green} = 1 - \Delta_\blue{\rm blue} - \pi(r_1^2 + 2r_3^2) \approx 0.133747976327$ . Therefore

$\large \boxed{A_\blue{\rm blue} > A_\green{\rm green}}$