Câu V:
$ab+bc+ca = 3 \rightarrow a+b+c \ge 3$
Xét: $\sum \dfrac{a}{1+b^2}=\sum (a-\dfrac{ab^2}{1+b^2}) \ge \sum (a-\dfrac{ab^2}
{2b}) = a+b+c-\dfrac{ab+bc+ca}{2}$
$\Longrightarrow \sum \dfrac{19a}{1+b^2} \ge 19(a+b+c)-\dfrac{19(ab+bc+ca)}{2}\
\bigstar$
Xét: $\sum \dfrac{1}{1+b^2} = \sum (1-\dfrac{b^2}{1+b^2}) \ge \sum (1-\dfrac{b^2}
{2b}) = \sum (1-\dfrac{b}{2})=3-\dfrac{a+b+c}{2}$
$\Longrightarrow \sum \dfrac{3}{1+b^2} \ge 9-\dfrac{3(a+b+c)}{2}\ \bigstar \bigstar$
Từ $\bigstar,\ \bigstar \bigstar$, ta có:
$T \ge 19(a+b+c)+9-\dfrac{19(ab+bc+ca)}{2}-\dfrac{3(a+b+c)}{2}$
$\iff T \ge \dfrac{35(a+b+c)}{2}-\dfrac{19.3}{2}+9$
$\iff T \ge \dfrac{35.3}{2}-\dfrac{57}{2}+9$
$\iff T \ge 33$
Dấu "=" xảy ra khi $a=b=c=1$