Quay lại đáp án

ta có: $\sum \frac{a^{2}}{a+2b^{3}}= \sum (a-\frac{2ab^{3}}{a+2b^{3}})\geqslant \sum (a-\frac{2ab^{3}}{3\sqrt[3]{ab^{6}}})= \sum (a-\frac{2}{3}\sqrt[3]{a^{2}}b)(1)$

ta có :$\sum \frac{a^{2}}{a+2b^{3}}= \sum (a-\frac{2ab^{3}}{a+2b^{3}})\geqslant \sum (a-\frac{2ab^{3}}{3\sqrt[3]{ab^{6}}})= \sum (a-\frac{2}{3}\sqrt[3]{a^{2}}b)(1)$Áp dụng BĐT cô si ta có :$\sum \sqrt[3]{a^{2}}b\leqslant \sum \frac{(a+a+1)b}{3}= \sum (\frac{2}{3}ab+\frac{1}{3}b)(2)$Từ (1)(2) ta có :$\sum (a-\frac{2}{3}\sqrt[3]{a^{2}}b)\geqslant \sum (a-\frac{4}{9}ab-\frac{2}{9}b)= \frac{7}{9}\sum a-\frac{4}{9}\sum ab\geqslant \frac{7}{9} \sum a-\frac{4}{27}(\sum a)^{2}= \frac{21}{9}-\frac{4}{27}.(3)^{2}=1$ xong