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Bổ đề: $\dfrac{1}{a}+\dfrac{1}{b} \ge \dfrac{4}{a+b}, \forall a, b >0$.
Áp dụng ta có
$\dfrac{16}{2x+y+z}=4.\dfrac{4}{2x+(y+z)} \le \dfrac{4}{2x}+\dfrac{4}{y+z} \le \dfrac{2}{x}+\dfrac{1}{y}+\dfrac{1}{z}$
Tương tự
$\dfrac{16}{x+2y+z}\le \dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z}$
$\dfrac{16}{x+y+2z}\le \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{2}{z}$
cộng theo từng vế ta có
$\dfrac{16}{2x+y+z}+\dfrac{16}{x+2y+z}+\dfrac{16}{x+y+2z} \le 4\left ( \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right )=16$
Vậy
$\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\leq 1$ , đpcm.
Dấu bằng xảy ra $\Leftrightarrow x=y=z=\dfrac{4}{3}$.
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