Question

Cho a, b, c > 0 thỏa mãn ab+bc+ca=3

CMR: a³+b³+c³ ≥ 3

Answer 1

Ta có: \(a^2+b^2+c^2\ge ab+bc+ca=3\)

Do \(a>0\Rightarrow\left(a-1\right)^2\left(a+\frac{1}{2}\right)\ge0\)

\(\Rightarrow a^3-\frac{3}{2}a^2+\frac{1}{2}\ge0\Rightarrow a^3\ge\frac{3}{2}a^2-\frac{1}{2}\)

Tương tự ta có: \(b^3\ge\frac{3}{2}b^2-\frac{1}{2}\) ; \(c^3\ge\frac{3}{2}c^2-\frac{1}{2}\)

Cộng vế với vế:

\(a^3+b^3+c^3\ge\frac{3}{2}\left(a^2+b^2+c^2\right)-\frac{3}{2}\ge3.\frac{3}{2}-\frac{3}{2}=3\)

Dấu "=" xảy ra khi \(a=b=c=1\)