11.[tex]p=a+b+c+\frac{1}{abc}=a+b+c+\frac{1}{9abc}+\frac{8}{9abc}\geq 4\sqrt[4]{a.b.c.\frac{1}{9abc}}+\frac{8}{9\sqrt{(\frac{a^2+b^2+c^2}{3})^3}}=..[/tex]
14.[tex]\sqrt[3]{a+b}=\sqrt[3]{\frac{9}{4}}.\sqrt[3]{a+b}.\sqrt[3]{\frac{2}{3}}.\sqrt[3]{\frac{2}{3}}\leq \sqrt[3]{\frac{9}{4}}.\frac{a+b+\frac{2}{3}+\frac{2}{3}}{3}=\sqrt[3]{\frac{1}{12}}.(a+b+\frac{4}{3})[/tex]
Đánh giá tương tự cho các hạng tử còn lại.
15.[tex]\sqrt[3]{a+2b}=\frac{1}{\sqrt[3]{9}}.\sqrt[3]{(a+2b).3.3}\leq \frac{1}{\sqrt[3]{9}}.\frac{a+2b+3+3}{3}[/tex]
Đánh giá tương tự cho các hạng tử còn lại.
10.[tex]S=\frac{1}{a^3+b^3}+\frac{1}{a^2b}+\frac{1}{ab^2}\geq \frac{1}{a^3+b^3}+\frac{4}{a^2b+ab^2}=\frac{1}{a^3+b^3}+\frac{1}{ab^2+a^2b}+\frac{1}{ab^2+a^2b}+\frac{1}{a^2b+ab^2}+\frac{1}{ab(a+b)}\geq \frac{16}{a^3+b^3+3a^2b+3ab^2}+\frac{1}{(a+b).\frac{(a+b)^2}{4}}=\frac{16}{(a+b)^3}+\frac{4}{(a+b)^3}+=\frac{20}{(a+b)^3}\geq 20[/tex]