We have to prove $\frac{1}{AX}+\frac{1}{XY}\geq \frac{4}{BC}$. After multiplication with BC, this inequality takes the form $\frac{BC}{AX}+\frac{BC}{XY}\geq 4$. But BC = BX + CX; thus, we can rewrite this inequality as $\frac{BX}{AX}+\frac{CX}{AX}+\frac{BX}{XY}+\frac{CX}{XY} \geq 4$. But this is clear, since $BX\cdot CX=AX\cdot XY$ by the Intersecting Chords Theorem, so that $\frac{BX\cdot CX}{AX\cdot XY}=1$, and thus AM-GM yields $\frac{BX}{AX}+\frac{CX}{AX}+\frac{BX}{XY}+\frac{CX}{XY} \geq 4\sqrt[4]{\frac{BX}{AX}\cdot\frac{CX}{AX}\cdot\frac{BX}{XY}\cdot\frac{CX}{XY}}$ $=4\sqrt[4]{\left(\frac{BX\cdot CX}{AX\cdot XY}\right)^2}=4\sqrt[4]{1^2}=4$.

Megus wrote: where can I find proof of Intersecting Chords Theorem, because to be honest, this is the first time I've come across it. You may have heard of it under the name Power of a Point. It's in AoPS vol. 1. Anyway, the proof for a point inside a circle is easy. Consider a point X inside a circle, and chords AC and BD passing through X. $\angle AXB$ and $\angle CXD$ are equal, due to the fact that they are vertical angles. Also, $\angle DCA$ and $\angle DBA$ are equal, since they are inscribed in the same arc. Thus, ABX and CDX are similar triangles. Therefore, we have $\frac{CX}{DX}=\frac{BX}{AX}$, or $CX\times AX=BX\times DX$, and we are done.

Let $AC,BD$ be intersecting chords, and let $X$ be their intersection point. Then, from the fact that $ABCD$ is cyclic, we get $\angle XAB=\angle XDC,\ \angle XBA=\angle XCD$, and we thus find the triangles $XAB,XDC$ to be similar. This means that $\frac{XA}{XB}=\frac{XD}{XC}$, which, in turn, proves $XA\cdot XC=XD\cdot XB$.

Holy **** After all I remarked what it is indeed - I've known this theorem well since few years... At least I know what does the chord [I cannot find it in a dictionary so far] mean eventually

Hm. I was just editing my post to include the proof... guess you beat me to it.