Không mất tính tổng quát, giả sử: $a\ge b\ge c \Rightarrow \dfrac{1}{3^a}\le\dfrac{1}{3^b}\le\dfrac{1}{3^c}$.
Ta có:
$(a-b)(\dfrac{1}{3^a}-\dfrac{1}{3^b})\le0 \Leftrightarrow \dfrac{a}{3^a}+\dfrac{b}{3^b}\le\dfrac{b}{3^a}+\dfrac{a}{3^b}$
$(a-c)(\dfrac{1}{3^a}-\dfrac{1}{3^c})\le0 \Leftrightarrow \dfrac{a}{3^a}+\dfrac{c}{3^c}\le\dfrac{c}{3^a}+\dfrac{a}{3^c}$
$(b-c)(\dfrac{1}{3^b}-\dfrac{1}{3^c})\le0 \Leftrightarrow \dfrac{b}{3^b}+\dfrac{c}{3^c}\le\dfrac{b}{3^c}+\dfrac{c}{3^b}$
Cộng 3 BĐT trên lại ta được:
$2(\dfrac{a}{3^a}+\dfrac{b}{3^b}+\dfrac{c}{3^c})\le\dfrac{b}{3^a}+\dfrac{a}{3^b}+\dfrac{c}{3^a}+\dfrac{a}{3^c}+\dfrac{b}{3^c}+\dfrac{c}{3^b}$
$\Leftrightarrow 3(\dfrac{a}{3^a}+\dfrac{b}{3^b}+\dfrac{c}{3^c})\le(a+b+c)(\dfrac{1}{3^a}+\dfrac{1}{3^b}+\dfrac{1}{3^c})$
$\Leftrightarrow 3(\dfrac{a}{3^a}+\dfrac{b}{3^b}+\dfrac{c}{3^c})\le\dfrac{1}{3^a}+\dfrac{1}{3^b}+\dfrac{1}{3^c}$
Dấu bằng xảy ra khi: $a=b=c=\dfrac{1}{3}$