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Có: $\frac{{bc}}{{{a^2}b + {a^2}c}} = \frac{{bc}}{{{a^2}\left( {b + c} \right)}} =
\frac{1}{{{a^2}\left( {\frac{1}{b} + \frac{1}{c}} \right)}} = \frac{{\frac{1}{{{a^2}}}}}{{\left(
{\frac{1}{b} + \frac{1}{c}} \right)}}$
Đặt $x = \frac{1}{a}\,;\,\,\,y = \frac{1}{b}\,\,\,;\,\,z = \frac{1}{c}$ thì giả thiết $a, b, c > 0; abc = 1$
$ \Leftrightarrow x,y,z > 0\,\,\,;\,\,xyz = 1\,;\,\,P = \frac{{{x^2}}}{{y + z}} + \frac{{{y^2}}}{{z +
x}} + \frac{{{z^2}}}{{y + x}}$
Theo bất đẳng thức Bunhiacopxki:
$\begin{array}{l}
(y + z + z + x + x + y).P \ge {\left( {\sqrt {y + z} \frac{x}{{\sqrt {y + z} }} + \sqrt {x + z}
\frac{y}{{\sqrt {x + z} }} + \sqrt {y + x} \frac{z}{{\sqrt {y + x} }}} \right)^2}\\
\Rightarrow 2(x + y + z).P \ge {(x + y + z)^2}\\
\Rightarrow P \ge \frac{1}{2}(x + y + z) \ge \frac{1}{2}.3\sqrt[3]{{xyz}} = \frac{1}{2}.3
\Rightarrow P \ge \frac{3}{2}
\end{array}$
Nếu $P = \frac{3}{2}$ thì $x = y = z = 1$ suy ra $a = b = c = 1$
Đảo lại, nếu $a = b = c = 1$ thì $P = \frac{3}{2}$.Vậy $\min P = \frac{3}{2}$
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