Question

Chứng minh \(\frac{a^3+b^3}{2}\ge\left(\frac{a+b}{2}\right)^3\)với a>0, b>0.

Answer 1

\(\frac{a^3+b^3}{2}\ge\left(\frac{a+b}{2}\right)^3\)

\(\Leftrightarrow\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{2}\ge\frac{\left(a+b\right)^3}{8}\)

\(\Leftrightarrow\frac{a^2-ab+b^2}{2}\ge\frac{\left(a+b\right)^2}{8}\)

\(\Leftrightarrow\frac{a^2-ab+b^2}{2}-\frac{a^2+2ab+b^2}{8}\ge0\)

\(\Leftrightarrow\frac{4a^2-4ab+4b^2-a^2-2ab-b^2}{8}\ge0\)

\(\Leftrightarrow\frac{3a^2-6ab+3b^2}{8}\ge0\)

\(\Leftrightarrow\frac{3\left(a-b\right)^2}{8}\ge0\) (luôn đúng)

Vậy \(\frac{a^3+b^3}{2}\ge\left(\frac{a+b}{2}\right)^3\)