Cho tứ diện P.ABC trong đó PA, PB, PC đôi một vuông góc.
Giải:
\begin{array}{l} Coi\,\left\{ \begin{array}{l} PA = m\\ PB = n\\ PC = p \end{array} \right. \Rightarrow \left\{ \begin{array}{l} {S_1} = {S_{PAB}} = \dfrac{{mn}}{2}\\ {S_2} = {S_{PBC}} = \dfrac{{np}}{2}\\ {S_3} = {S_{PCA}} = \dfrac{{pm}}{2} \end{array} \right.\& S = {S_{ABC}} = \dfrac{{AH.BC}}{2} = \dfrac{{\sqrt {{n^2} + {p^2}} }}{2}.\,\sqrt {P{A^2} + P{H^2}} \\ = \dfrac{{\sqrt {{n^2} + {p^2}} }}{2}.\sqrt {{m^2} + \dfrac{{{n^2}{p^2}}}{{{n^2} + {p^2}}}} = \dfrac{1}{2}\sqrt {{{\left( {mn} \right)}^2} + {{\left( {np} \right)}^2} + {{\left( {pm} \right)}^2}} \Leftrightarrow {S^2} = \dfrac{1}{4}\left( {4S_1^2 + 4S_2^2 + 4S_3^2} \right)\\ \Leftrightarrow {S^2} = S_1^2 + S_2^2 + S_3^2 \Rightarrow M = \dfrac{{S_1^2}}{{2S_1^2 + S_2^2 + S_3^2}} + \dfrac{{S_2^2}}{{S_1^2 + 2S_2^2 + S_3^2}} + \dfrac{{S_3^2}}{{S_1^2 + S_2^2 + 2S_3^2}}\\ Coi\,\left\{ \begin{array}{l} S_1^2 = a\\ S_2^2 = b\\ S_3^2 = c \end{array} \right. \Rightarrow M = \dfrac{a}{{2a + b + c}} + \dfrac{b}{{a + 2b + c}} + \dfrac{c}{{a + b + 2c}}\left( {a,b,c > 0} \right)\\ Coi\,\left\{ \begin{array}{l} x = 2a + b + c\\ y = a + 2b + c\\ z = a + b + 2c \end{array} \right. \Rightarrow \left\{ \begin{array}{l} a = \dfrac{{3x - y - z}}{4}\\ b = \dfrac{{ - x + 3y - z}}{4}\\ c = \dfrac{{ - x - y + 3z}}{4} \end{array} \right. \Rightarrow M = \dfrac{1}{4}\left( {3 - \dfrac{y}{x} - \dfrac{z}{x} + 3 - \dfrac{x}{y} - \dfrac{z}{y} + 3 - \dfrac{x}{z} - \dfrac{y}{z}} \right)\\ \Rightarrow M = \dfrac{9}{4} - \dfrac{1}{4}\left[ {\left( {\dfrac{x}{y} + \dfrac{y}{x}} \right) + \left( {\dfrac{y}{z} + \dfrac{z}{y}} \right) + \left( {\dfrac{x}{z} + \dfrac{z}{x}} \right)} \right] \le \dfrac{9}{4} - \dfrac{1}{4}\left( {2 + 2 + 2} \right) = \dfrac{9}{4} - \dfrac{3}{2} = \dfrac{3}{4}\\ \Rightarrow Max\,M = \dfrac{3}{4} \Leftrightarrow x = y = z \Leftrightarrow a = b = c \Leftrightarrow {S_1} = {S_2} = {S_3} = \dfrac{S}{{\sqrt 3 }} \Leftrightarrow m = n = p \end{array}
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