Cho x,y,z>0 thỏa mãn x+y+z =1 Tìm GTNN của: M=(x+1/x)^2+(y+1/y)^2+(z+1/z)^2

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Cho x,y,z>0 thỏa mãn x+y+z =1
Tìm GTNN của: M=(x+1/x)^2+(y+1/y)^2+(z+1/z)^2

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Emery 2 tháng 2021-07-18T20:28:59+00:00 1 Answers 0 views 0

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    0
    2021-07-18T20:30:25+00:00

    Đáp án:

    \[\frac{{100}}{3}\]

    Giải thích các bước giải:

     Ta có:

    \(\begin{array}{l}
    {\left( {x – y} \right)^2} \ge 0 \Rightarrow {x^2} – 2xy + {y^2} \ge 0 \Rightarrow {x^2} + {y^2} \ge 2xy\\
    {\left( {y – z} \right)^2} \ge 0 \Rightarrow {y^2} – 2yz + {z^2} \ge 0 \Rightarrow {y^2} + {z^2} \ge 2yz\\
    {\left( {z – x} \right)^2} \ge 0 \Rightarrow {z^2} – 2zx + {x^2} \ge 0 \Rightarrow {z^2} + {x^2} \ge 2zx\\
     \Rightarrow \left( {{x^2} + {y^2}} \right) + \left( {{y^2} + {z^2}} \right) + \left( {{z^2} + {x^2}} \right) \ge 2xy + 2yz + 2zx\\
     \Rightarrow {x^2} + {y^2} + {z^2} \ge xy + yz + zx\\
    x + y + z = 1\\
     \Leftrightarrow {x^2} + {y^2} + {z^2} + 2xy + 2yz + 2zx = 1\\
     \Leftrightarrow 1 \le {x^2} + {y^2} + {z^2} + 2\left( {{x^2} + {y^2} + {z^2}} \right) = 3\left( {{x^2} + {y^2} + {z^2}} \right)\\
     \Rightarrow {x^2} + {y^2} + {z^2} \ge \frac{1}{3}\\
    1 = x + y + z \ge 3.\sqrt[3]{{x.y.z}} \Rightarrow xyz \le \frac{1}{{27}}\\
    M = {\left( {x + \frac{1}{x}} \right)^2} + {\left( {y + \frac{1}{y}} \right)^2} + {\left( {z + \frac{1}{z}} \right)^2}\\
     = {x^2} + 2x.\frac{1}{x} + \frac{1}{{{x^2}}} + {y^2} + 2.y.\frac{1}{y} + \frac{1}{{{y^2}}} + {z^2} + 2.z.\frac{1}{z} + \frac{1}{{{z^2}}}\\
     = \left( {{x^2} + {y^2} + {z^2}} \right) + \left( {\frac{1}{{{x^2}}} + \frac{1}{{{y^2}}} + \frac{1}{{{z^2}}}} \right) + 6\\
     \ge \frac{1}{3} + 3.\sqrt[3]{{\frac{1}{{{x^2}}}.\frac{1}{{{y^2}}}.\frac{1}{{{z^2}}}}} + 6\\
     \ge \frac{1}{3} + 3.\sqrt[3]{{\frac{1}{{\frac{1}{{{{27}^2}}}}}}} + 6 = \frac{1}{3} + 27 + 6 = \frac{{100}}{3}
    \end{array}\)

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