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\[\begin{array}{l}
*{a^2} \ge {a^2} - {(b - c)^2} \\
\Leftrightarrow {a^2} \ge (a - b + c)(a + b - c) \\
*{b^2} \ge {b^2} - {(a - c)^2} \\
\Leftrightarrow {a^2} \ge (b - a + c)(b + a - c) \\
*{c^2} \ge {c^2} - {(a - b)^2} \\
\Leftrightarrow {c^2} \ge (c - a + b)(c + a - b) \\
\\
\end{array}\]
Suy ra:

\[\begin{array}{l}
{a^2}{b^2}{c^2} \ge {(b - a + c)^2}{(b + a - c)^2}{(a - b + c)^2} \\
\Leftrightarrow abc \ge (a - b + c)(a + b - c)(b - a + c)(dpcm) \\
\end{array}\]

1)Cho $\left\{ \begin{array}{l}
x,\,y,\,z > 0\\
\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 4
\end{array} \right.$ Tìm GTNN của $P = \frac{1}{{2x + y + z}} + \frac{1}{{2y + x + z}} + \frac{1}{{2z + x + y}}$
*Chứng minh:
$\frac{1}{4}.\left( {\frac{1}{a} + \frac{1}{b}} \right) \ge \frac{1}{{a + b}}$
*Có:
${a^2} - 2ab + {b^2} \ge 0\forall \,a,\,b$
$\begin{array}{l}
 \Leftrightarrow {a^2} + {b^2} \ge 2ab\\
 \Leftrightarrow {a^2} + 2ab + {b^2} \ge 4ab
\end{array}$
$ \Leftrightarrow \frac{{{{\left( {a + b} \right)}^2}}}{{ab}} \ge 4$
$ \Leftrightarrow \frac{1}{a} + \frac{1}{b} \ge \frac{4}{{a + b}}$
$ \Leftrightarrow \frac{1}{4}.\left( {\frac{1}{a} + \frac{1}{b}} \right) \ge \frac{1}{{a + b}}$
Đẳng thức xảy ra khi $a=b$
Áp dụng vào bài ta được:
$\frac{1}{{x + y}} \le \frac{1}{4}.\left( {\frac{1}{x} + \frac{1}{y}} \right)$ (1)
$\frac{1}{{x + z}} \le \frac{1}{4}.\left( {\frac{1}{x} + \frac{1}{z}} \right)$ (2)
Cộng (1), (2) ta được:
$\frac{1}{{x + y}} + \frac{1}{{x + z}} \le \frac{1}{4}.\left( {\frac{2}{x} + \frac{1}{y} + \frac{1}{z}} \right)$
$ \Leftrightarrow \frac{1}{4}.\left( {\frac{1}{{x + y}} + \frac{1}{{x + z}}} \right) \le \frac{1}{{16}}.\left( {\frac{2}{x} + \frac{1}{y} + \frac{1}{z}} \right)$
Mà $\frac{1}{4}.\left( {\frac{1}{{x + y}} + \frac{1}{{x + z}}} \right) \ge \frac{1}{{2x + y + z}}$
Suy ra $\frac{1}{{2x + y + z}} \le \frac{1}{{16}}.\left( {\frac{2}{x} + \frac{1}{y} + \frac{1}{z}} \right)$ (3)
Đẳng thức xảy ra khi $x = y = z = \frac{3}{4}$
*Có:
$\begin{array}{l}
\frac{1}{{y + z}} \le \frac{1}{4}.\left( {\frac{1}{y} + \frac{1}{z}} \right)\left( 4 \right)\\
\frac{1}{{x + y}} \le \frac{1}{4}.\left( {\frac{1}{x} + \frac{1}{y}} \right)\left( 1 \right)
\end{array}$
Cộng (1), (4) ta được:
$\frac{1}{{y + z}} + \frac{1}{{x + y}} \le \frac{1}{4}.\left( {\frac{1}{x} + \frac{2}{y} + \frac{1}{z}} \right)$
$ \Leftrightarrow \frac{1}{4}.\left( {\frac{1}{{y + z}} + \frac{1}{{x + y}}} \right) \le \frac{1}{{16}}.\left( {\frac{1}{x} + \frac{2}{y} + \frac{1}{z}} \right)$
Mà $\frac{1}{{2y + x + z}} \le \frac{1}{4}.\left( {\frac{1}{{y + z}} + \frac{1}{{x + y}}} \right)$
Suy ra $\frac{1}{{2y + x + z}} \le \frac{1}{{16}}.\left( {\frac{1}{x} + \frac{2}{y} + \frac{1}{z}} \right)$
Đẳng thức xảy ra $x = y = z = \frac{3}{4}$ (5)
*Có:
$\frac{1}{{y + z}} \le \frac{1}{4}.\left( {\frac{1}{y} + \frac{1}{z}} \right)\left( 4 \right)$
$\frac{1}{{x + z}} \le \frac{1}{4}.\left( {\frac{1}{x} + \frac{1}{z}} \right)\left( 2 \right)$
Cộng (2), (4) ta được:
$\frac{1}{{x + z}} + \frac{1}{{y + z}} \le \frac{1}{4}.\left( {\frac{1}{x} + \frac{1}{y} + \frac{2}{z}} \right)$
$ \Leftrightarrow \frac{1}{4}.\left( {\frac{1}{{x + z}} + \frac{1}{{y + z}}} \right) \le \frac{1}{{16}}.\left( {\frac{1}{x} + \frac{1}{y} + \frac{2}{z}} \right)$
Mà $\frac{1}{4}.\left( {\frac{1}{{x + z}} + \frac{1}{{y + z}}} \right) \ge \frac{1}{{x + y + 2z}}$
Suy ra $\frac{1}{{x + y + 2z}} \le \frac{1}{{16}}.\left( {\frac{1}{x} + \frac{1}{y} + \frac{2}{z}} \right)\left( 6 \right)$
Đẳng thức xảy ra $x = y = z = \frac{3}{4}$ (5)
*Cộng (3), (5), (6) ta được:
$\frac{1}{{2x + y + z}} + \frac{1}{{2y + x + z}} + \frac{1}{{2z + x + y}} \le \frac{1}{{16}}.\left( {\frac{4}{x} + \frac{4}{y} + \frac{4}{z}} \right)$
$ \Leftrightarrow \frac{1}{{2x + y + z}} + \frac{1}{{2y + x + z}} + \frac{1}{{2z + x + y}} \le \frac{1}{{16}}.16 = 1$
Đẳng thức xay ra khi $x = y = z = \frac{3}{4}$
Vậy $Max\,P = 1\,khi\,x = y = z = \frac{3}{4}$