Một cách tìm Min, Max lạ mà hay!

Đề bài: (Câu hỏi của bạn Mã Mã hỏi trên facebook Trợ Giúp Toán Học)\[Cho\,\left\{ \begin{array}{l}
x,y,z \in \left[ {1;4} \right]\\
x \ge y\& x \ge z
\end{array} \right..\,Min\,P = \frac{x}{{2x + 3y}} + \frac{y}{{y + z}} + \frac{z}{{z + x}} = ?\]Giải:\
[\begin{array}{l}
P = f\left( z \right) = \frac{x}{{2x + 3y}} + \frac{y}{{y + z}} + \frac{z}{{z + x}} = 1 - \frac{x}{{z + x}} + \frac{y}{{y + z}} + \frac{x}{{2x + 3y}}\\
 \Rightarrow f'\left( z \right) = \frac{x}{{{{\left( {z + x} \right)}^2}}} - \frac{y}{{{{\left( {z + y} \right)}^2}}} = \frac{{{{\left( {z\sqrt x  + y\sqrt x } \right)}^2} - {{\left( {z\sqrt y  + x\sqrt y } \right)}^2}}}{{{{\left[ {\left( {z + x} \right)\left( {z + y} \right)} \right]}^2}}}\\
 = \frac{{\left( {z\sqrt x  + y\sqrt x  + z\sqrt y  + x\sqrt y } \right)\left( {z\sqrt x  - y\sqrt x  - z\sqrt y  + x\sqrt y } \right)}}{{{{\left[ {\left( {z + x} \right)\left( {z + y} \right)} \right]}^2}}}\\
 = \frac{{\left( {z\sqrt x  + y\sqrt x  + z\sqrt y  + x\sqrt y } \right)\left[ {z\left( {\sqrt x  - \sqrt y } \right) - \sqrt {xy} \left( {\sqrt x  - \sqrt y } \right)} \right]}}{{{{\left[ {\left( {z + x} \right)\left( {z + y} \right)} \right]}^2}}}\\
 = \frac{{\left( {z\sqrt x  + y\sqrt x  + z\sqrt y  + x\sqrt y } \right)\left( {\sqrt x  - \sqrt y } \right)\left( {z - \sqrt {xy} } \right)}}{{{{\left[ {\left( {z + x} \right)\left( {z + y} \right)} \right]}^2}}}\\
Do\,\left\{ \begin{array}{l}
z\sqrt x  + y\sqrt x  + z\sqrt y  + x\sqrt y  > 0\\
\sqrt x  - \sqrt y  \ge 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
f'\left( z \right) > 0 \Leftrightarrow z > \sqrt {xy} \\
f'\left( z \right) < 0 \Leftrightarrow z < \sqrt {xy}
\end{array} \right.\\
 \Rightarrow P \ge \mathop {Min}\limits_{\left[ {1;4} \right]} \,f\left( z \right) = f\left( {\sqrt {xy} } \right) = \frac{x}{{2x + 3y}} + \frac{y}{{y + \sqrt {xy} }} + \frac{{\sqrt {xy} }}{{\sqrt {xy}  + x}}\\
 = \frac{{\frac{x}{y}}}{{2\frac{x}{y} + 3}} + \frac{1}{{1 + \sqrt {\frac{x}{y}} }} + \frac{1}{{1 + \sqrt {\frac{x}{y}} }} = \frac{{{t^2}}}{{2{t^2} + 3}} + \frac{2}{{1 + t}} = f\left( t \right)\,\left( {1 = \sqrt {\frac{x}{x}}  \le t = \sqrt {\frac{x}{y}}  \le \sqrt x  \le \sqrt 4  = 2} \right)\\
 \Rightarrow f'\left( t \right) = \frac{{6t}}{{{{\left( {2{t^2} + 3} \right)}^2}}} - \frac{2}{{{{\left( {t + 1} \right)}^2}}} = \frac{{ - 2\left[ {4{t^3}\left( {t - 1} \right) + 3\left( {2{t^2} - t + 3} \right)} \right]}}{{{{\left( {2{t^2} + 3} \right)}^2}{{\left( {t + 1} \right)}^2}}} < 0,\forall t \in \left[ {1;2} \right]\\
 \Rightarrow P \ge \mathop {Min}\limits_{\left[ {1;2} \right]} \,f\left( t \right) = f\left( 2 \right) = \frac{{34}}{{33}} \Rightarrow Min\,P = \frac{{34}}{{33}} \Leftrightarrow \left\{ \begin{array}{l}
z = \sqrt {xy} \\
\sqrt {\frac{x}{y}}  = 2\\
x = 4
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = 4\\
y = 1\\
z = 2
\end{array} \right.
\end{array}\]Sau đây là một số bài tập tương tự có sử dụng phương pháp này:\[\begin{array}{l}
 \bullet \,Cho\,a,b,c \in \left[ {\frac{1}{3};3} \right].\, \to Max\,P = \frac{a}{{a + b}} + \frac{b}{{b + c}} + \frac{c}{{c + a}} = ?\\
 \bullet \,Cho\,\left\{ \begin{array}{l}
a,b,c \in R\\
abc + a + c = b
\end{array} \right..\,Max\,P = \frac{2}{{{a^2} + 1}} - \frac{2}{{{b^2} + 1}} + \frac{3}{{{c^2} + 1}} = ?\\
 \bullet \,Cho\left\{ \begin{array}{l}
a,b,c > 0\\
21ab + 2bc + 8ca \le 12
\end{array} \right..\,Min\,P = \frac{1}{a} + \frac{2}{b} + \frac{3}{c} = ?\\
 \bullet \,Cho\,\left\{ \begin{array}{l}
a,b,c \ge 0\\
a + b + c = 1
\end{array} \right..\,Max\,M = 3\left( {{a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2}} \right) + 3\left( {ab + bc + ca} \right) + 2\sqrt {{a^2} + {b^2} + {c^2}} \\
 \bullet \,Cho\,\left\{ \begin{array}{l}
x,y,z > 0\\
x + y + z = 4\\
xyz = 3
\end{array} \right..\,CMR:183 - 165\sqrt 5  \le {x^4} + {y^4} + {z^4} \le 18
\end{array}\]