Question

Chứng minh rằng: 3a³+7b³\(\ge\)9ab², với \(\forall\)a,b\(\ge\)0

Answer 1

\(3a^3+7b^3\ge3a^3+6b^3\)

\(=3a^3+3b^3+3b^3\)

\(\ge3\sqrt[3]{3.a^3.3.b^3.3.b^3}=9ab^2\)

Dấu = xảy ra khi a = b = 0

Answer 2

\(3a^3+\frac{7}{2}b^3+\frac{7}{2}b^3\ge3\sqrt[3]{3a^3.\frac{7}{2}b^3.\frac{7}{2}b^3}=ab^2.3\sqrt[3]{\frac{147}{4}}>9ab^2\)