Bài giải của hocmai.toanhoc ( Trịnh Hào Quang)
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Đề bài:
Cho [TEX]\left\{ \begin{array}{l}a,b,c > 0 \\\frac{1}{{a + 2}} + \frac{2}{{b + 4}} \le \frac{{c + 2}}{{c + 5}} \\\end{array} \right.[/TEX]
Tìm Min: [TEX]P = (a + 1)(b + 2)(c + 2)[/TEX]
Giải:
Đặt:
[TEX]\left\{ \begin{array}{l}x = a + 1 \\y = b + 2 \\z = c + 2 \\\end{array} \right. \Rightarrow \left\{ \begin{array}{l}x > 1;y > 2;z > 2 \\\frac{1}{{x + 1}} + \frac{1}{{y + 2}} \le \frac{z}{{z + 3}} \\\end{array} \right.\,\,;P = xyz[/TEX]
[TEX]\begin{array}{l}\frac{{2\sqrt 2 }}{{\left( {x + 1} \right)\left( {y + 2} \right)}} \le \frac{1}{{x + 1}} + \frac{2}{{y + 2}} \le \frac{z}{{z + 3}} \Rightarrow z \ge \frac{{2\sqrt 2 \left( {z + 3} \right)}}{{\sqrt {\left( {x + 1} \right)\left( {y + 2} \right)} }}\\Do:1 + \frac{1}{{x + 1}} + \frac{1}{{y + 2}} \le \frac{z}{{z + 3}} + 1 \Leftrightarrow 1 - \frac{z}{{z + 3}} + \frac{1}{{x + 1}} \le + 1 - \frac{1}{{y + 2}} \Leftrightarrow \frac{3}{{z + 3}} + \frac{1}{{x + 1}} \le \frac{y}{{y + 2}} \\\frac{{2\sqrt 3 }}{{\left( {x + 1} \right)\left( {z + 3} \right)}} \le \frac{3}{{z + 3}} + \frac{1}{{x + 1}} \le \frac{y}{{y + 2}} \Rightarrow y \ge \frac{{2\sqrt 3 \left( {y + 2} \right)}}{{\sqrt {\left( {x + 1} \right)\left( {z + 3} \right)} }} \\\& \,\frac{{2\sqrt 6 }}{{\left( {y + 2} \right)\left( {z + 3} \right)}} \le \frac{2}{{y + 2}} + \frac{3}{{z + 3}} \le \frac{x}{{x + 1}} \Leftrightarrow x \ge \frac{{2\sqrt 6 \left( {x + 1} \right)}}{{\sqrt {\left( {y + 2} \right)\left( {z + 3} \right)} }} \\P = xyz \ge \frac{{2\sqrt 2 \left( {z + 3} \right)}}{{\sqrt {\left( {x + 1} \right)\left( {y + 2} \right)}}}.\frac{{2\sqrt 3 \left( {y + 2} \right)}}{{\sqrt {\left( {x + 1} \right)\left( {z + 3} \right)} }}.\frac{{2\sqrt 6 \left( {x + 1} \right)}}{{\sqrt {\left( {y + 2} \right)\left( {z + 3} \right)} }} = 8.6 = 48 \Rightarrow MinP = 48. \\\left\{ \begin{array}{l}\frac{1}{{x + 1}} = \frac{2}{{y + 2}} = \frac{3}{{z + 3}} = t \\xyz = 48 \\\end{array} \right. \Rightarrow \left\{ \begin{array}{l}x = t - 1 \\y = 2(t - 1) \\z = 3(t - 1) \\6(t - 1)^3 = 48 \\\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 2 \\y = 4 \\z = 6 \\\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 1 \\b = 2 \\c = 4 \\\end{array} \right.\\\\\end{array}[/TEX]
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