Xét hiệu \(\left(a^3+b^3\right)-\frac{1}{4}\left(a+b\right)^3\) ta có:
\(\left(a^3+b^3\right)-\frac{1}{4}\left(a+b\right)^3=\frac{1}{4}\left[4\left(a^3+b^3\right)-\left(a+b\right)^3\right]\)
\(=\frac{1}{4}\left[4a^3+4b^3-\left(a^3+3a^2b+3ab^2+b^3\right)\right]\)\(=\frac{1}{4}\left(4a^3+4b^3-a^3-3a^2b-3ab^2-b^3\right)\)
\(=\frac{1}{4}\left(3a^3+3b^3-3a^2b-3ab^2\right)\)\(=\frac{3}{4}\left(a^3+b^3-a^2b-ab^2\right)\)
\(=\frac{3}{4}\left[\left(a^3-a^2b\right)+\left(b^3-ab^2\right)\right]\)\(=\frac{3}{4}\left[a^2\left(a-b\right)+b^2\left(b-a\right)\right]\)
\(=\frac{3}{4}\left[a^2\left(a-b\right)-b^2\left(a-b\right)\right]\)\(=\frac{3}{4}\left(a-b\right)\left(a^2-b^2\right)\)\(=\frac{3}{4}\left(a-b\right)^2\left(a+b\right)\)
Vì a và b > 0 \(\Rightarrow a+b>0\)
mà \(\left(a-b\right)^2\ge0\forall a,b\)và \(\frac{3}{4}>0\)
\(\Rightarrow\frac{3}{4}\left(a-b\right)^2\left(a+b\right)\ge0\)
hay \(\left(a^3+b^3\right)-\frac{1}{4}\left(a+b\right)^3\ge0\)\(\Rightarrow a^3+b^3\ge\frac{1}{4}\left(a+b\right)^3\)
