Junior Balkan Mathematical Olympiad

Solution by PSC. At first, we will allow $P$ and $Q$ to coincide, and we will prove the inequality in this case. Let the perpendicular from $Q$ to $O Q$ meet $\Omega$ at $B$ and $C$. Then, we have that $Q B=Q C$. We will show that
$$B Q \cdot X Y \geq 2 Q X \cdot Q Y$$
By the power of a point Theorem we have that
$$Q X \cdot Q Y=Q B \cdot Q C=Q B^{2}$$
therefore it is enough to prove that $X Y \geq 2 B Q$ or $X Y \geq B C$. Let $T$ be the foot of the perpendicular from $O$ to $X Y$. Then, from the right-angled triangle $O T Q$ we have that $O T \leq O Q$, so the distance from $O$ to the chord $X Y$ is smaller or equal to the distance from $O$ to the chord $B C$. This means that $X Y \geq B C$, so (1) holds.

Back to the initial problem, we have to prove that
$$M P \cdot X Y \geq 2 Q X \cdot P Y \Longleftrightarrow \frac{X Y}{2 Q X} \geq \frac{P Y}{P M}$$
By (1) we have that
$$\frac{X Y}{2 Q X} \geq \frac{Q Y}{Q B}$$
so it is enough to prove that
$$\frac{Q Y}{Q B} \geq \frac{P Y}{P M}$$
If $C B$ meets $Y M$ at $S$, then from $M P \| Q S$ we get
$$\frac{Q Y}{Q B} \geq \frac{Q Y}{Q S}=\frac{P Y}{P M}$$
which is the desired.

We will show that $(Q M-Q P) \cdot X Y \geq 2 \cdot Q X \cdot P Y$. Since $M P \geq Q M-Q P$, our inequality follows directly. Let $A$ the intersection point of $\ell$ with the diameter which passes through $Q$. Like in the following picture, choose a coordinative system centered at $O$ and such that $Q=(a, 0), A=(c, 0)$, $P=(c, h)$ and denote the lengths $Q X=x, P Q=t, P Y=y, O P=d, Q M=z$.
Let $\lambda_{Q}=r^{2}-a^{2}$ and $\lambda_{P}=r^{2}-d^{2}$ respectively the power of $Q$ and $P$ with respect to our circle $\Omega$. We will show that:
$$(z-t)(t+x+y) \geq 2 x y$$

Adding and multiplying respectively the relations $x(t+y)=\lambda_{Q}$ and $y(t+x)=\lambda_{P}$, we will have
$$t(x+y)+2 x y=\lambda_{P}+\lambda_{Q}$$
and
$$x y(t+x)(t+y)=\lambda_{P} \lambda_{Q}$$
Using these two equations, it's easy to deduce that:
$$(x y)^{2}-x y\left(t^{2}+\lambda_{P}+\lambda_{Q}\right)+\lambda_{P} \lambda_{Q}=0$$
So, $w_{1}=x y$ is a zero of the second degree polynomial:
$$p(w)=w^{2}-w\left(t^{2}+\lambda_{P}+\lambda_{Q}\right)+\lambda_{P} \lambda_{Q}$$
But $w_{1}=x y<x(t+y)=\lambda_{Q}$ and
$$\begin{aligned} p\left(\lambda_{Q}\right) & =\left(r^{2}-a^{2}\right)^{2}-\left(r^{2}-a^{2}\right)\left(t^{2}+\lambda_{P}+\lambda_{Q}\right)+\lambda_{P} \lambda_{Q} \\ & =\left(r^{2}-a^{2}\right)^{2}-\left(r^{2}-a^{2}\right)\left(t^{2}+r^{2}-d^{2}+r^{2}-a^{2}\right)+\left(r^{2}-d^{2}\right)\left(r^{2}-a^{2}\right) \\ & =\left(r^{2}-a^{2}\right)^{2}-\left(r^{2}-a^{2}\right) t^{2}-\left(r^{2}-d^{2}\right)\left(r^{2}-a^{2}\right)-\left(r^{2}-a^{2}\right)^{4}+\left(r^{2}-d^{2}\right)\left(r^{2}-a^{2}\right) \\ & =-t^{2}\left(r^{2}-a^{2}\right)=-t^{2} \lambda_{Q}<0 \end{aligned}$$
This implies that $\lambda_{Q}$ lies (strictly) between the two (positive) zeros $w_{1}, w_{2}$ of $p(w)$ and $w_{1}=x y$ is the smaller one.
After using (2) and (3), inequality (1) can be rewritten as:
$$(x y)^{2} \leq\left(\frac{z-t}{z+t}\right) \lambda_{P} \lambda_{Q}$$
In order to show this, it is enough to show that
$$p\left(\sqrt{\frac{z-t}{z+t} \lambda_{P} \lambda_{Q}}\right) \leq 0$$
because this will imply $\sqrt{\frac{z-t}{z+t} \lambda_{P} \lambda_{Q}} \in\left[w_{1}, w_{2}\right]$. After some manipulations, inequality (6) can be equivalently transformed to:
$$4 z^{2} \lambda_{P} \lambda_{Q} \leq\left(z^{2}-t^{2}\right)\left(t^{2}+\lambda_{P}+\lambda_{Q}\right)^{2}$$
Since $z^{2}-t^{2}=r^{2}-d^{2}=\lambda_{P}$, this is equivalent to:
$$4 z^{2} \lambda_{Q} \leq\left(t^{2}+\lambda_{P}+\lambda_{Q}\right)^{2}$$
But $t^{2}=(a-c)^{2}+h^{2}=a^{2}+d^{2}-2 a c, z^{2}=t^{2}+r^{2}-d^{2}=a^{2}-2 a c+r^{2}$ and $t^{2}+\lambda_{P}+\lambda_{Q}=\ldots=2\left(r^{2}-a c\right)$. Hence, (8) is equivalent with:
$$\left(a^{2}-2 a c+r^{2}\right)\left(r^{2}-a^{2}\right) \leq\left(r^{2}-a c\right)^{2} \Leftrightarrow \cdots \Leftrightarrow 0 \leq a^{2}(a-c)^{2}$$
which is clearly true.