Giải thích các bước giải:
Áp dụng bất đẳng thức $\dfrac1x+\dfrac1y\ge\dfrac4{x+y}$
$\to \dfrac1{x+y}\le\dfrac14(\dfrac1x+\dfrac1y)$
Ta có:
$\dfrac1{a+2b+c}=\dfrac1{(a+b)+(b+c)}\le \dfrac14(\dfrac1{a+b}+\dfrac1{b+c})$
$\to \dfrac1{a+2b+c}\le \dfrac14(\dfrac14(\dfrac1a+\dfrac1b)+\dfrac14(\dfrac1b+\dfrac1c))$
$\to\dfrac1{a+2b+c}\le \dfrac1{16a}+\dfrac1{8b}+\dfrac1{16c}(1)$
Tương tự chứng minh được:
$\dfrac1{b+2c+a}\le \dfrac1{16b}+\dfrac1{8c}+\dfrac1{16c}(2)$
$\dfrac1{c+2a+b}\le\dfrac1{16c}+\dfrac1{8a}+\dfrac1{16b}(3)$
Cộng vế với vế của $(1),(2),(3)$ ta có:
$\dfrac1{a+2b+c}+\dfrac1{b+2c+a}+\dfrac1{c+2a+b}\le \dfrac14(\dfrac1a+\dfrac1b+\dfrac1c)=1$
$\to đpcm$