cho x,y,z >0 thoả mãn 1/x+1/y+1/z=4 c/m:A=1/(2x+y+z)+1/(x+2y+z)+1/x+y+2z

04/12/2021 Bởi Arya

0 bình luận về “cho x,y,z >0 thoả mãn 1/x+1/y+1/z=4 c/m:A=1/(2x+y+z)+1/(x+2y+z)+1/x+y+2z<hoặc =1 Giúp mình với,mình cần gấp!!!”

1. Đáp án + giải thích các bước giải:

   Áp dụng bất đẳng thức cộng mẫu, ta có:

   $\left\{\begin{matrix} \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x}+\dfrac{1}{z}\ge\dfrac{(1+1+1+1)^2}{x+y+x+z}=\dfrac{16}{2x+y+z}\\\dfrac{1}{y}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{(1+1+1+1)^2}{y+x+y+z}=\dfrac{16}{x+2y+z}\\ \dfrac{1}{z}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{x}\ge\dfrac{(1+1+1+1)^2}{z+y+z+x}=\dfrac{16}{x+y+2z} \end{matrix}\right. \\ \to \left\{\begin{matrix} \dfrac{1}{2x+y+z}\le\dfrac{1}{16}(\dfrac{2}{x}+\dfrac{1}{y}+\dfrac{1}{z})\\\dfrac{1}{x+2y+z}\le\dfrac{1}{16}(\dfrac{2}{y}+\dfrac{1}{x}+\dfrac{1}{z})\\ \dfrac{1}{x+y+2z}\le\dfrac{1}{16}(\dfrac{2}{z}+\dfrac{1}{x}+\dfrac{1}{y}) \end{matrix}\right. \\ \to \dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\le\dfrac{1}{16}(\dfrac{2}{x}+\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{2}{z}+\dfrac{1}{z}+\dfrac{1}{z})\\\to \dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\le \dfrac{1}{16}(\dfrac{4}{x}+\dfrac{4}{y}+\dfrac{4}{z})\\\to \dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\le\dfrac{1}{4}(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z})\\\to \dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\le\dfrac{1}{4}.4=1\\\to đpcm$

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