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

Đáp án:

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