Cho $x,y,z>0$ thỏa mãn điều kiện $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=4$ CMR: $\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\leq1$

Ta chứng minh $\dfrac{1}{a}+\dfrac{1}{b}\ge \dfrac{4}{a+b}$

BĐT tương đương: $\dfrac{a+b}{ab}\ge \dfrac{4}{a+b}$

$\Leftrightarrow (a+b)^2\ge  4ab$

$\Leftrightarrow (a-b)^2\ge 0$(luôn đúng)

Áp dụng ta có: 

$\dfrac{1}{2x+y+z}=\dfrac{1}{x+y+x+z}\le \dfrac{1}{4}(\dfrac{1}{x+y}+\dfrac{1}{x+z})\le \dfrac{1}{16}(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x}+\dfrac{1}{z})$

Chứng minh tương tự: $\dfrac{1}{x+2y+z}\le \dfrac{1}{16}(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z});\dfrac{1}{x+y+2z}\le \dfrac{1}{16}(\dfrac{1}{x}+\dfrac{1}{z}+\dfrac{1}{y}+\dfrac{1}{z})$

Cộng vế theo vế ta được:

$\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\\ \le \dfrac{1}{16}(\dfrac{2}{x}+\dfrac{1}{y}+\dfrac{1}{z})+\dfrac{1}{16}(\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z})+\dfrac{1}{16}(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{2}{z})\\ \le \dfrac{1}{16}(\dfrac{2}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{2}{z})\\ \le \dfrac{1}{16}(\dfrac{4}{x}+\dfrac{4}{y}+\dfrac{4}{z})=\dfrac{1}{16}.4(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z})=\dfrac{1}{4}.4=1(ĐPCM)$

Đẳng thức xảy ra khi $x=y=z=\dfrac{3}{4}$