It's well-known that $\triangle{XPQ}\sim\triangle{XBC}$ and $\triangle{XBP}\sim\triangle{XCQ}$ (all similarities here are direct). First we prove some easy configuration facts: $\angle{XPQ}=\angle{XBC}$ forces $X,A$ to lie the same side of $BC$, whence $\angle{PXB}=\angle{QXC}$ forces $X,A$ to lie on the same side of $PQ$. Let $\alpha=\angle{PXB}=\angle{QXC}$ and $\beta=\angle{XBP}=\angle{XCQ}$ so that $\alpha<\beta<90^\circ$ (since $\triangle{ABC}$ is acute and $PX>PB$). For convenience, define the similarity ratio $r = PQ/BC = XP/XB = \sin\beta/\sin(\alpha+\beta)$. For points $U,V,W$, let $d(U,VW)$ denote the directed distance from $U$ to $VW$ (directed so that $X,A$ are "above" $PQ$ and $BC$). Then $d(Y,BC) = d(X,BC) - XY\cos\alpha$ and $XY = 2\cdot d(X,PQ) = 2r\cdot d(X,BC)$. It follows that \begin{align*} \frac{2[YBC]}{2[XPQ]} = \frac{2[YBC]}{2r^2[XBC]} = \frac{| BC\cdot d(Y,BC)|}{r^2 BC\cdot d(X,BC)} &= \frac{|d(X,BC) - 2r\cdot d(X,BC)\cos\alpha|}{r^2\cdot d(X,BC)} \\ &= \frac{\sin(\alpha+\beta)}{\sin^2\beta}|\sin(\alpha+\beta)-2\sin\beta\cos\alpha| \\ &= \frac{|\sin(\alpha+\beta)\sin(\alpha-\beta)|}{\sin^2\beta} \\ &= \frac{|\sin^2\alpha - \sin^2\beta|}{\sin^2\beta}<1, \end{align*}as desired.

Interesting problem Let $ PQ $ and $ BC $ meet at $ T $. And circle $ (TXY) $ meet $ PQ,BC $ at $ R,S $ (1) First, we show that $ Y $ is below $ BC $. (Actually, $ Y $ is on $ BC $ when $ PX=PB $) $ R $ is the midpoint of arc $ XY $. And $ \triangle{XRS} \sim \triangle{XPB} \sim \triangle{XQC} $. Which means $ XR >RS $ so, $ Y $ is below $ BC $. (2) Now we show the problem. $ \frac{S_{\bigtriangleup YPQ}}{S_{\bigtriangleup YBC}}= \frac{{PQ} \times {RY}}{{BC} \times {SY}}=\frac{{XR} \times {RY}}{{XS} \times {SY}} >1 $. So, done