Bài 8: $VT=\sum \frac{a(b+c)}{a^2+(b+c)^2}=\sum \frac{a(b+c)}{a^2+\frac{1}{4}(b+c)^2+\frac{3}{4}(b+c)^2}\leq \sum \frac{a(b+c)}{a(b+c)+\frac{3}{4}(b+c)^2}=\sum \frac{a}{a+\frac{3}{4}(b+c)}$
Đặt $P=(3-VT).\frac{4}{3}=\sum \frac{b+c}{a+\frac{3}{4}(b+c)}$
Và $S=\sum (b+c).\left [ a+\frac{3}{4}(b+c) \right ]=\frac{3}{2}(\sum a^2)+\frac{7}{2}(\sum ab)=\frac{3}{2}(\sum a)^2+\frac{1}{2}(\sum ab)\leq 1.5(\sum a)^2+\frac{1}{6}(\sum a)^2=\frac{5}{3}(\sum a)^2$
Áp dụng bđt Holder tc: $P.S\geq (b+c+a+b+c+a)^2=4(\sum a)^2$
$\Rightarrow P\geq \frac{12}{5}\Rightarrow (3-VT).\frac{4}{3}\geq \frac{12}{5}\Rightarrow VT\leq \frac{6}{5}(đpcm)$
Dấu ''='' xr khi a=b=c>0