[TIKZ]\draw[thick] (0,0) circle (3cm);
\coordinate[label=left:$P_1$] (A) at (-1.96,-2.28);
\coordinate[label=left:$P_7$] (B) at (-3,0);
\coordinate[label=left:$P_6$] (C) at (-1.94,2.3);
\coordinate[label=above:$P_5$] (D) at (0.54,2.92);
\coordinate[label=right:$P_4$] (E) at (2.5,1.7);
\coordinate[label=right:$P_3$] (F) at (2.88,-0.9);
\coordinate[label=below:$P_2$] (G) at (0.7,-2.9);
\coordinate[label=below:$a$] (H) at (-0.4,-2.26);
\coordinate[label=below:$b$] (I) at (1.2,-0.86);
\coordinate[label=below:$c$] (J) at (0.7,0.6);
\draw (A) -- (B) -- (C) -- (D) -- (E) -- (F)--(G)--(A);
\draw (A) -- (D);
\draw (A) -- (F);
\draw (A) -- (E);
\draw (D) -- (F);
[/TIKZ]
Let $P_1P_2=a,\,P_1P_3=b,\,P_1P_4=c$. By the Ptolomeus thoerem for the quadrangle $P_1P_3P_4P_5$ it follows that $ab+ac=bc$, i.e. $\dfrac{a}{b}+\dfrac{a}{c}=1$. Since $\triangle P_1P_2P_3\equiv \triangle Q_1Q_2Q_3$, then $\dfrac{Q_1Q_2}{Q_1Q_3}=\dfrac{a}{b}$ and hence $Q_1Q_2=\dfrac{a^2}{b}$.
Analogously $R_1R_2=\dfrac{a^2}{c}$. Therefore, $\dfrac{S_Q+S_R}{S_P}=\dfrac{a^2}{b}+\dfrac{a^2}{c}$. Then $\dfrac{a^2}{b}+\dfrac{a^2}{c}>\dfrac{1}{2}\left(\dfrac{a}{b}+\dfrac{a}{c}\right)^2=\dfrac{1}{2}$ (equality is not possible because $\dfrac{a}{b}\ne\dfrac{a}{c}$.
On the other hand
$\dfrac{a^2}{b^2}+\dfrac{a^2}{c^2}=\left(\dfrac{a}{b}+\dfrac{a}{c}\right)^2-\dfrac{2a^2}{bc}=1-\dfrac{2a^2}{bc}$---(1)
By the Sine theorem, we get
$\dfrac{a^2}{bc}=\dfrac{\sin^2 \dfrac{\pi}{7}}{\sin^2 \dfrac{\pi}{7}\sin^2 \dfrac{4\pi}{7}}=\dfrac{1}{4\cos\dfrac{2\pi}{7}\left(1+\cos\dfrac{2\pi}{7}\right)}$
Since $\cos\dfrac{2\pi}{7}<\cos \dfrac{\pi}{4}=\dfrac{\sqrt{2}}{2}$, then $\dfrac{a^2}{bc}>\dfrac{1}{4\dfrac{\sqrt{2}}{2}\left(1+\dfrac{\sqrt{2}}{2}\right)}=\sqrt{2}-1$. From here and from (1), we get the right hand side inequality of the problem.