câu 1:
$\sum \frac{a^{3}}{(b+1)(c+1)}= \sum \frac{a^{4}}{a(abc+b)(abc+c)}= \sum \frac{a^{4}}{(bc+1)(ac+1)}\geq \sum \frac{4a^{4}}{(ac+bc+2)^{2}}\geq \frac{4}{3}(\sum \frac{a^{2}}{ab+ac+2})^{2}(1)$
áp dụng bđt schwars ta có
$\sum \frac{a^{2}}{ab+ac+2}\geq \frac{(a+b+c)^{2}}{2(ab+ac+bc+3)}$
vì
$(a+b+c)^{2}\geq 9\sqrt[3]{(abc)^{2}}=9$
$ab+bc+ac\leq \frac{1}{3}(a+b+c)^{2}$
nên
$\frac{(a+b+c)^{2}}{2(ab+bc+ac+3)}\geq \frac{3}{4}$
suy ra $\sum \frac{a^{2}}{ab+ac+2}\geq \frac{3}{4}(2)$
từ (1)(2) ta được đpcm