Giải thích các bước giải:
Áp dụng bđt svacxo ta có
$\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge \dfrac{(1+1+1+1)^2}{a+a+b+c}=\dfrac{16}{2a+b+c}$
$\rightarrow \dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{16}{2a+b+c} (*)$
Tương tự ta có:
$\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{1}{c}\ge\dfrac{16}{a+2b+c} (**)$
$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{c}\ge\dfrac{16}{a+b+2c} (***)$
Cộng vế với vế (*),(**),(***) ta được
$16(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c})\le 4(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})$
$\rightarrow (\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c})\le \dfrac{1}{4}(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})$
