cho 1/x+1/y+1/z =8 tìm GTLN của P = 1/(2x+y+z) +1/(x+2y+z) + 1/(x+y+2z)

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1. Đáp án:

   Giải thích các bước giải:

   $\begin{array}{l}
   Su\,dung\,\frac{1}{x} + \frac{1}{y} \ge \frac{4}{{x + y}}\\
   Ta\,co:\frac{1}{{2x + y + z}} = \frac{1}{4}.\frac{4}{{\left( {x + y} \right) + \left( {x + z} \right)}}\\
   \le \frac{1}{4}\left( {\frac{1}{{x + y}} + \frac{1}{{x + z}}} \right) = \frac{1}{{16}}\left( {\frac{4}{{x + y}} + \frac{4}{{x + z}}} \right)\\
   \le \frac{1}{{16}}\left( {\frac{1}{x} + \frac{1}{y} + \frac{1}{x} + \frac{1}{z}} \right) = \frac{1}{{16}}\left( {\frac{2}{x} + \frac{1}{y} + \frac{1}{z}} \right)\\
   Tuong\,tu\,\frac{1}{{x + 2y + z}} \le \frac{1}{{16}}\left( {\frac{1}{x} + \frac{2}{y} + \frac{1}{z}} \right)\\
   \frac{1}{{x + y + 2z}} \le \frac{1}{{16}}\left( {\frac{1}{x} + \frac{1}{y} + \frac{2}{z}} \right)\\
   \Rightarrow \frac{1}{{2x + y + z}} + \frac{1}{{x + 2y + z}} + \frac{1}{{x + y + 2z}} \le \frac{1}{{16}}\left( {\frac{4}{x} + \frac{4}{y} + \frac{4}{z}} \right)\\
   = \frac{1}{4}\left( {\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right) = \frac{1}{4}.8 = 2\\
   Dau\, = \,xay\,ra\,khi\,x = y = z = \frac{3}{8}
   \end{array}$

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