The answer is in the affirmative: Given a positive $\epsilonY_{\alpha} Y_{\pi+\alpha}$ yields $2 \cdot \frac{1}{2}\left(f^{2}(\alpha)+f^{2}(\pi+\alpha)\right)=O X_{\alpha}^{2}+O X_{\pi+\alpha}^{2} \geq \frac{1}{2} X_{\alpha} X_{\pi+\alpha}^{2}>$ $\frac{1}{2} Y_{\alpha} Y_{\pi+\alpha}^{2} \geq \frac{1}{2}\left(O Y_{\alpha}^{2}+O Y_{\pi+\alpha}^{2}\right)=\frac{1}{2}\left(g^{2}(\alpha)+g^{2}(\pi+\alpha)\right)$. Integration then gives us $2[P]>\left[P^{\prime}\right]$, as needed.

This can also be proved via elementary methods. Actually, we will establish the following more general fact.

Fact. Let $P=A_{1} A_{2} A_{3} A_{4}$ and $P^{\prime}=B_{1} B_{2} B_{3} B_{4}$ be two convex quadrangles in the plane, and let $O$ be one of their common points different from the vertices of $P^{\prime}$. Denote by $\ell_{i}$ the line $O B_{i}$, and assume that for every $i=1,2,3,4$ the length of segment $\ell_{i} \cap P$ is greater than the length of segment $\ell_{i} \cap P^{\prime}$. Then $\left[P^{\prime}\right]\left[P^{\prime}\right]\left(1-\frac{B_{1} C_{3}+B_{4} C_{2}}{2 B_{1} B_{4}}\right) \geq \frac{\left[P^{\prime}\right]}{2} .
\end{aligned}
A contradiction. Case 2. Assume now that the rays $B_{1} B_{2}$ and $B_{4} B_{3}$ intersect at some point (see the right figure above). Denote by $L$ the common point of $B_{2} C_{1}$ and $B_{3} C_{4}$. We have $\left[B_{2} C_{4} C_{1}\right] \geq\left[B_{2} C_{4} B_{3}\right]$, hence $\left[C_{1} C_{4} L\right] \geq\left[B_{2} B_{3} L\right]$. Thus we have
$$\begin{aligned} {\left[P^{\prime}\right]>\left[B_{1} B_{2} C_{1} C_{2}\right]+\left[B_{3} B_{4} C_{3} C_{4}\right] } & =\left[P^{\prime}\right]+\left[L C_{1} C_{2} C_{3} C_{4}\right]-\left[B_{2} B_{3} L\right] \\ & \geq\left[P^{\prime}\right]+\left[C_{1} C_{4} L\right]-\left[B_{2} B_{3} L\right] \geq\left[P^{\prime}\right] . \end{aligned}$$
$$$$

A final contradiction.