Question

Cho \(a,b>0\) . CMR: \(\frac{a^3+b^3}{2}\ge\left(\frac{a+b}{2}\right)^3\)

Answer 1

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

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

\(\Leftrightarrow8\left(a^3+b^3\right)\ge2\left(a^3+3a^2b+3ab^2+b^3\right)\)

\(\Leftrightarrow4a^3+4b^3-a^3-3a^2b-3ab^2-b^3\ge0\)

\(\Leftrightarrow3a^3-3a^2b-3ab^2+3b^3\ge0\)

\(\Leftrightarrow a^3-a^2b-ab^2+b^3\ge0\)

\(\Leftrightarrow a^2\left(a-b\right)-b^2\left(a-b\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2-b^2\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\)

( Luôn đúng với mọi \(a;b>0\) )

Dấu "=" xảy ra \(\Leftrightarrow a=b\)