Relation Involving Lengths from Vertex to Circumcircle Through Triangle Side

To determine the correct relation, we analyze the geometric properties of the triangle and its circumcircle.

Step 1: Power of a Point (Intersecting Chords Theorem)

Let the circumcircle of $\triangle PQR$ be denoted by $\Gamma$.
The line segment $PT$ is a chord of $\Gamma$ passing through the vertex $P$ and intersecting the side $QR$ (which is another chord of $\Gamma$) at the point $S$.
Since the two chords $PT$ and $QR$ intersect internally at $S$, by the Intersecting Chords Theorem, the products of their segments are equal:
$PS \times ST = QS \times SR$

Step 2: Applying the AM-GM Inequality to $PS$ and $ST$

Consider the positive real quantities $\frac{1}{PS}$ and $\frac{1}{ST}$. By the Arithmetic Mean - Geometric Mean (AM-GM) inequality:
$\frac{\frac{1}{PS} + \frac{1}{ST}}{2} \geq \sqrt{\frac{1}{PS} \times \frac{1}{ST}}$

Since $S$ is not the centre of the circumcircle, the segments $PS$ and $ST$ are not equal in general ($PS \neq ST$), which gives the strict inequality:
$\frac{1}{PS} + \frac{1}{ST} > \frac{2}{\sqrt{PS \times ST}}$

Substituting $PS \times ST = QS \times SR$ into the inequality, we get:
$\frac{1}{PS} + \frac{1}{ST} > \frac{2}{\sqrt{QS \times SR}}$
This establishes that option (B) is correct.

Step 3: Applying the AM-GM Inequality to $QS$ and $SR$

Similarly, applying the AM-GM inequality to the segments $QS$ and $SR$:
$\frac{QS + SR}{2} \geq \sqrt{QS \times SR}$

Since $S$ is not the centre of the circumcircle, $S$ does not bisect the chord $QR$ while simultaneously bisecting $PT$, meaning $QS \neq SR$. Thus:
$\sqrt{QS \times SR} < \frac{QS + SR}{2} = \frac{QR}{2}$

Taking the reciprocal on both sides reverses the inequality:
$\frac{1}{\sqrt{QS \times SR}} > \frac{2}{QR}$

Multiplying both sides by $2$:
$\frac{2}{\sqrt{QS \times SR}} > \frac{4}{QR}$

Step 4: Combining the Inequalities

From Step 2 and Step 3, we have:
$\frac{1}{PS} + \frac{1}{ST} > \frac{2}{\sqrt{QS \times SR}} > \frac{4}{QR}$

Therefore:
$\frac{1}{PS} + \frac{1}{ST} > \frac{4}{QR}$
This establishes that option (D) is also correct.

Conclusion

The correct options are B and D.