SAMC

(a) Assume that the origin of the coordinates system is at $A$. Let $r$ be the inradius of $△ABC$, and $I$ the incenter. Then $I(r,r)$ and the line $d$ has equation $$d:y-r=m(x-r)$$ that is $y=r+m(x-r)$. We get $$P\left(\frac{r(m-1)}{m},0\right),Q(0,r(m-1)).$$ It follows $$\begin{array}{rl}b\cdot \frac{PB}{PA}& +c\cdot \frac{QC}{QA}=b\cdot \frac{c-\frac{r(m-1)}{m}}{\frac{r(m-1)}{m}}+c\cdot \frac{b+r(m-1)}{-r(m-1)}\\ & =b\cdot \frac{mc-r(m-1)}{r(m-1)}+c\cdot \frac{b+r(m-1)}{-r(m-1)}\\ & =b\cdot \frac{mc}{r(m-1)}-c\cdot \frac{b}{r(m-1)}-(b+c)\\ & =\frac{bc}{r(m-1)}(m-1)-(b+c)=\frac{bc}{r}-(b+c).\end{array}$$ But $r=\frac{2K}{a+b+c}=\frac{bc}{a+b+c}$, hence $\frac{bc}{r}=a+b+c$, and we get: $$b\cdot \frac{PB}{PA}+c\cdot \frac{QC}{QA}=(a+b+c)-(b+c)=a.$$

(b) From the previous relation we have $$\frac{b}{a}\cdot \frac{PB}{PA}+\frac{c}{a}\cdot \frac{QC}{QA}=1,$$ so from Cauchy-Schwarz inequality it follows $$\left[{\left(\frac{PB}{PA}\right)}^2+{\left(\frac{QC}{QA}\right)}^2\right]\left(\frac{b^2}{a^2}+\frac{c^2}{a^2}\right)\ge {\left(\frac{b}{a}\cdot \frac{PB}{PA}+\frac{c}{a}\cdot \frac{QC}{QA}\right)}^2=1$$ hence $${\left(\frac{PB}{PA}\right)}^2+{\left(\frac{QC}{QA}\right)}^2\ge 1$$ The minimum value is $1$ and it is obtained for the line $d$ satisfying the property $$\frac{PB}{PA}\cdot \frac{a}{b}=\frac{QC}{QA}\cdot \frac{a}{c}.$$

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Alternative solution.

(a) (Abdullah Al-Saeed). The relation is equivalent to $$b\cdot \frac{c-PA}{PA}+c\cdot \frac{b-QA}{QA}=a$$ that is $$\begin{array}{c}bc\left(\frac{1}{PA}+\frac{1}{QA}\right)=a+b+c\end{array}\text{\hspace{1em}(1)}$$ We know that $K[ABC]=\frac{bc}{2}=sr$, where $s$ is the semiperimeter of triangle $ABC$, hence $$r=\frac{bc}{a+b+c},$$ and by replacing in (1) we obtain $$\begin{array}{c}\frac{r}{PA}+\frac{r}{QA}=1\end{array}\text{\hspace{1em}(2)}$$ From the similarity we have $\frac{r}{PA}=\frac{QI}{QP}$ and $\frac{r}{QA}=\frac{PI}{QP}$, and a relation (2) follows.

(b) From the previous relation we have $$\frac{b}{a}\cdot \frac{PB}{PA}+\frac{c}{a}\cdot \frac{QC}{QA}=1,$$ so from Cauchy-Schwarz inequality it follows $$\left[{\left(\frac{PB}{PA}\right)}^2+{\left(\frac{QC}{QA}\right)}^2\right]\left(\frac{b^2}{a^2}+\frac{c^2}{a^2}\right)\ge {\left(\frac{b}{a}\cdot \frac{PB}{PA}+\frac{c}{a}\cdot \frac{QC}{QA}\right)}^2=1$$ hence $${\left(\frac{PB}{PA}\right)}^2+{\left(\frac{QC}{QA}\right)}^2\ge 1$$ The minimum value is $1$ and it is obtained for the line $d$ satisfying the property $$\frac{PB}{PA}\cdot \frac{a}{b}=\frac{QC}{QA}\cdot \frac{a}{c}.$$