Lời giải:$\sum \dfrac{a(b+c)}{a^{2}+(b+c)^{2}}=\sum \dfrac{a(b+c)}{a^{2}+\dfrac{1}{4}(b+c)^{2}+\dfrac{3}{4}(b+c)^{2}}\leq \sum \dfrac{a(b+c)}{a(b+c)+\frac{3}{4}(b+c)^{2}}$
Ta có: $1-\dfrac{a(b+c)}{a(b+c)+\dfrac{3}{4}(b+c)^{2}}$
$=\dfrac{3}{4}\cdot \dfrac{(b+c)^{2}}{a(b+c)+\dfrac{3}{4}(b+c)^{2}}$
$\Rightarrow 3-\sum \dfrac{a(b+c)}{a(b+c)+\dfrac{3}{4}(b+c)^{2}}$
$=\dfrac{3}{4}\sum \dfrac{(b+c)^{2}}{a(b+c)+\dfrac{3}{4}(b+c)^{2}}$
$\geq \dfrac{3}{4}\cdot \dfrac{4(a+b+c)^{2}}{\dfrac{3}{2}(a^{2}+b^{2}+c^{2})+\dfrac{7}{2}(ab+bc+ca)}$
$= 6\cdot \dfrac{(a+b+c)^{2}}{3(a+b+c)^{2}+ab+bc+ca}\geq 6\cdot \dfrac{(a+b+c)^{2}}{3(a+b+c)^{2}+\dfrac{(a+b+c)^{2}}{3}}$
$= \dfrac{9}{5}\Rightarrow \sum \dfrac{a(b+c)}{a(b+c)+\frac{3}{4}(b+c)^{2}}\leq \dfrac{6}{5}$
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