Giải thích các bước giải:
Ta có:
$\dfrac1a+\dfrac1a+\dfrac1b+\dfrac1c\ge \dfrac{(1+1+1+1)^2}{a+a+b+c}=\dfrac{16}{2a+b+c}$
$\to \dfrac2a+\dfrac1b+\dfrac1c\ge \dfrac{16}{2a+b+c}(1)$
Tương tự:
$\dfrac1a+\dfrac2b+\dfrac1c\ge\dfrac{16}{a+2b+c}(2)$
$\dfrac1a+\dfrac1b+\dfrac2c\ge\dfrac{16}{a+b+2c}(3)$
Cộng vế với vế của $(1), (2), (3)$
$\to 16(\dfrac1{2a+b+c}+\dfrac1{a+2b+c}+\dfrac1{a+b+2c})\le 4(\dfrac1a+\dfrac1b+\dfrac1c)=16$
$\to \dfrac1{2a+b+c}+\dfrac1{a+2b+c}+\dfrac1{a+b+2c}\le 1$
$\to đpcm$