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Let \(BC=a, CA=b, AB=c, s=(a+b+c)/2\). Then by AM-GM \[36r^2=\frac{36(s-a)(s-b)(s-c)}{s}\le \frac{4s^2}{3}=\frac{(a+b+c)^2}{3}\le a^2+b^2+c^2=2(m^2+n^2+p^2).\]Thus (a) is proved. As for (b) let \(H_1\) be the foot of perpendicular from \(S\) to \(BC\). The upper bound follows from the triangle inequality: \[\frac{r_1}{H}=\frac{H_1A}{AD+DH_1+H_1A}<\frac{1}{2}.\]Now the lower bound. WLOG \(DH_1=1\). Then \(AH_1=\sqrt{m^2+1}\). We wish to find the largest \(\alpha\) for which \[\frac{r_1}{H}=\frac{\sqrt{m^2+1}}{\sqrt{m^2+1}+m+1}\ge \alpha \quad \iff \quad(2\alpha-\alpha^2)m^2+2m+(2\alpha-\alpha^2)\ge 0.\]It will be true for any real \(m\) if its discriminant wrt \(m\) is nonpositive: \[(\alpha-1)^2((\alpha+1)^2-2)\le 0 \quad \iff\quad \alpha \le \sqrt 2 -1.\]Thus \(\displaystyle \frac{r_1}{H}\ge \sqrt 2-1>0.4.\)