In the figure prove that $$\min(OR,PR,QR)+OR+PR+QR<OQ+QP+PO$$
This is a stronger form of the inequality $OR+PR+QR<OQ+QP+PO$ that can be easily proven.I have made some constructions in the above figure.From that we can write $$ \frac{RU}{OU}+\frac{RS}{QS}+\frac{RT}{PT} =1$$ I also have tried to use the fact $$RU<\max(RP,RQ),SR<\max(OR,RP) ...$$
Any ideas
Say we have an arbitrary point $R$ in the interior of the triangle $\triangle OPQ$.
WLOG, $\angle PSQ \gt \angle OSQ$, then $\triangle PSQ$ is an obtuse angled triangle and $\angle OSQ \lt 90^0$ (or vice versa). We draw a perp from point $O$ to segment $QS$ which meets it at point $V$.
We also note that if $R$ is between point $S$ and $V$, $\angle ORQ \lt 90^0$ but $3$ angles at point $R \, (\angle ORQ, \angle PRQ, \angle PRO)$ make $360^0$. So at least two of those angles will be $ \gt 90^0$ and we pick those two favorable one's for our construct. In this case we are assuming those are $\angle ORQ$ and $\angle PRQ$.
So we have $PR + QR \lt (PS + SR) + QR = PS + SQ \lt PS + QP$ ...(i)
Now, $QR + OR \lt QR + (RV + VO) = QV + VO \lt OQ + OS$ ...(ii)
If $QS$ is perpendicular to $OP$, both inequalities still hold.
So from (i) + (ii), we have
$OR + PR + QR + QR \lt QP + (PS + OS) + OQ = OQ + QP + PO$
$OR + PR + QR + min(OR,PR, QR) \lt OQ + QP + PO$