Ta có : $3a^2+2b^2+c^2$
$ = 2.(a^2+b^2)+a^2+c^2$
$≥ 2.2ab+2ac$
$ = 4ab + 2ac$
$\to \dfrac{a}{3a^2+2b^2+c^2} ≤ \dfrac{a}{4ab + 2ac} = \dfrac{1}{4b+2c} = \dfrac{1}{36}.\dfrac{6^2}{4b+2c}$
$≤ \dfrac{1}{36}.\bigg(\dfrac{4}{b} + \dfrac{2}{c}\bigg)$ ( BĐT Svac - xơ )
Chứng minh tương tự ta có :
$\dfrac{b}{3b^2+2c^2+a^2} ≤ \dfrac{1}{36}.\bigg(\dfrac{4}{c} + \dfrac{2}{a}\bigg)$
$\dfrac{c}{3c^2+2a^2+b^2} ≤ \dfrac{1}{36}.\bigg(\dfrac{4}{a} + \dfrac{2}{b}\bigg)$
Cộng vế với vế các BĐT trên ta được :
$\dfrac{a}{3a^2+2b^2+c^2} + \dfrac{b}{3b^2+2c^2+a^2} + \dfrac{c}{3c^2+2a^2+b^2}$
$≤ \dfrac{1}{6}\bigg(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\bigg)$
