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Since the function $\cos x$ is concave for $0 \le x \le \tfrac12\pi$ , the AM/GM inequality tells us that $\sqrt[3]{\cos\tfrac12A \cos\tfrac12B \cos\tfrac12C} \; \le \; \tfrac13\big(\cos\tfrac12A + \cos\tfrac12B + \cos\tfrac12C\big) \; \le \; \cos\big(\tfrac{A+B+C}{6}\big) \; = \; \tfrac12\sqrt{3}$ and so $\cos\tfrac12A \cos\tfrac12B \cos\tfrac12C \; \le \; \tfrac38\sqrt{3}$ Letting $s$ represent the semiperimeter, this tells us that $\sqrt{\frac{s(s-a)}{bc} \times \frac{s(s-b)}{ac} \times \frac{s(s-c)}{ab}} \; \le \; \tfrac38\sqrt{3}$ and hence $s\Delta \; \le \; \tfrac38\sqrt{3}abc$ where $\Delta$ is the area of the triangle. Equality occurs in all the above inequalities when the triangle is equilateral.

Standard triangle formulae tell us that (here $r$ is the inradius) $AI \; = \; \sqrt{(s-a)^2 + r^2} \; = \; \sqrt{(s-a)^2 + \frac{(s-a)(s-b)(s-c)}{s}} \; = \; \sqrt{\frac{(s-a)bc}{s}}$ and, similarly, $BI \; = \; \sqrt{\frac{(s-b)ac}{s}} \hspace{2cm} CI \; = \; \sqrt{\frac{(s-c)ab}{s}}$ Note that $\angle DBI \; = \; \angle DBC + \angle CBI \;=\; \tfrac12A + \tfrac12B \; = \; 90^\circ - \tfrac12C$ while $\angle IDB = C$ . Hence the triangle $DBI$ is isosceles, and so $DI \; = \; DB \; = \; \frac{IB}{2\sin \frac12C} \; = \; \tfrac12a\sqrt{\frac{bc}{s(s-a)}}$ and hence $AD \; = \; \sqrt{\frac{(s-a)bc}{s}} + \tfrac12a\sqrt{\frac{bc}{s(s-a)}} \; = \; \tfrac12\sqrt{\frac{bc}{s(s-a)}}\big[2(s-a) + a\big] \; = \; \tfrac12\sqrt{\frac{bc}{s(s-a)}}(b+c)$ and, similarly, $BE \; = \; \tfrac12\sqrt{\frac{ac}{s(s-b)}}(a+c) \hspace{2cm} CF \; = \; \tfrac12\sqrt{\frac{ab}{s(s-c)}}(a+b)$ Since $\Delta = \tfrac12ah_a \; = \; \tfrac12bh_b \; = \; \tfrac12ch_c$ , we deduce that $P \; = \; AD h_a + BE h_b + CF h_c \; = \; \frac{b+c}{a}\sqrt{bc(s-b)(s-c)} + \frac{a+c}{b}\sqrt{ac(s-a)(s-c)} + \frac{a+b}{c}\sqrt{ab(s-a)(s-b)}$ Splitting $\frac{b+c}{a} = \frac{b}{a} + \frac{c}{a}$ and so on (treating $P$ as a sum of six terms), the AM/GM inequality gives $\begin{aligned} P & \ge 6\Big[\frac{bc}{a^2} bc(s-b)(s-c) \times \frac{ac}{b^2} ac(s-a)(s-b) \times \frac{ab}{c^2} ab(s-a)(s-b)\Big]^{\frac16} \\ & = 6\Big[a^2b^2c^2(s-a)^2(s-b)^2(s-c)^2\Big]^{\frac16} \; = \; 6\Big[abc(s-a)(s-b)(s-c)\Big]^{\frac13} \\ & = 6\Big[\frac{abc\Delta^2}{s}\Big]^{\frac13} \; \ge \; 6\Big[\frac{8\Delta^3}{3\sqrt{3}}\Big]^{\frac13} \; = \; 4\sqrt{3}\Delta \end{aligned}$ with equality when the triangle is equilateral. In this case, with $\Delta = 1$ , we see that the minimum value of $P$ is $\boxed{4\sqrt{3}}$ .

Form a cyclic expression in (a,b,c) . Minimize using AM-GM