parmenides51 wrote: Let $XY$ be a chord of a circle $\Omega$, with center $O$, which is not a diameter. Let $P, Q$ be two distinct points inside the segment $XY$, where $Q$ lies between $P$ and $X$. Let $\ell$ the perpendicular line dropped from $P$ to the diameter which passes through $Q$. Let $M$ be the intersection point of $\ell$ and $\Omega$, which is closer to $P$. Prove that $$ MP \cdot XY \ge 2 \cdot QX \cdot PY$$ Let $\ell’$ be the line through $Q$ parallel to $\ell$. Let $U, V = \ell’ \cap \Omega$ and $R = \ell’ \cap PM$. Claim : $XY \cdot QU \geq 2 \cdot QX \cdot QY$ Proof : Note that $UV \leq XY$ as $XY$ is farther from the centre than $UV$. Also, $Q$ is midpoint of $UV$. So, $XY \cdot QU \geq 2 \cdot QU^2 = 2 \cdot QX \cdot QY$. $\square$ Since $QR \geq QU$, we have $\frac{XY}{2 \cdot QX} \geq \frac{QY}{QU} \geq \frac{QY}{QR} = \frac{PY}{PM}$ which gives the desired result. $\blacksquare$

MathDelicacy12 wrote: parmenides51 wrote: Let $XY$ be a chord of a circle $\Omega$, with center $O$, which is not a diameter. Let $P, Q$ be two distinct points inside the segment $XY$, where $Q$ lies between $P$ and $X$. Let $\ell$ the perpendicular line dropped from $P$ to the diameter which passes through $Q$. Let $M$ be the intersection point of $\ell$ and $\Omega$, which is closer to $P$. Prove that $$ MP \cdot XY \ge 2 \cdot QX \cdot PY$$ Let $\ell’$ be the line through $Q$ parallel to $\ell$. Let $U, V = \ell’ \cap \Omega$ and $R = \ell’ \cap YM$. Claim : $XY \cdot QU \geq 2 \cdot QX \cdot QY$ Proof : Note that $UV \leq XY$ as $XY$ is closer to the centre than $UV$. Also, $Q$ is midpoint of $UV$. So, $XY \cdot QU \geq 2 \cdot QU^2 = 2 \cdot QX \cdot QY$. $\square$ Since $QR \geq QU$, we have $\frac{XY}{2 \cdot QX} \geq \frac{QY}{QU} \geq \frac{QY}{QR} = \frac{PY}{PM}$ which gives the desired result. $\blacksquare$ I think you meant this.