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Giải thích các bước giải:
Ta có :
$\dfrac{a}{b+c}+\dfrac{b+c}{a}+\dfrac{b}{a+c}+\dfrac{c+a}{b}+\dfrac{c}{a+b}+\dfrac{a+b}{c}$
$=(\dfrac{a}{b+c}+\dfrac{b+c}{4a})+(\dfrac{b}{a+c}+\dfrac{c+a}{4b})+(\dfrac{c}{a+b}+\dfrac{a+b}{4c})+\dfrac{3}{4}.(\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c})$
$=(\dfrac{a}{b+c}+\dfrac{b+c}{4a})+(\dfrac{b}{a+c}+\dfrac{c+a}{4b})+(\dfrac{c}{a+b}+\dfrac{a+b}{4c})+\dfrac{3}{4}.(\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{c}{b}+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{c})$
$\ge 2\sqrt{\dfrac{a}{b+c}.\dfrac{b+c}{4a}}+2\sqrt{\dfrac{b}{a+c}.\dfrac{c+a}{4b}}+2\sqrt{\dfrac{c}{a+b}.\dfrac{a+b}{4c}}+\dfrac{3}{4}.6\sqrt[6]{\dfrac{b}{a}.\dfrac{c}{a}.\dfrac{c}{b}.\dfrac{a}{b}.\dfrac{a}{c}.\dfrac{b}{c}}$
$\ge \dfrac{15}{2}$
$\rightarrow đpcm$
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