Question

Chứng minh bất phương trình : a² + \(\dfrac{b^2}{4}\)\(\ge\)ab

Answer 1

\(BDT\Leftrightarrow a^2+\dfrac{b^2}{4}-ab\ge0\)

\(\Leftrightarrow\dfrac{1}{4}\left(4a^2+b^2-4ab\right)\ge0\)

\(\Leftrightarrow\dfrac{1}{4}\left[4a^2-2ab-2ab+b^2\right]\ge0\)

\(\Leftrightarrow\dfrac{1}{4}\left[2a\left(2a-b\right)-b\left(2a-b\right)\right]\ge0\)

\(\Leftrightarrow\dfrac{1}{4}\left(2a-b\right)\left(2a-b\right)\ge0\)

\(\Leftrightarrow\dfrac{1}{4}\left(2a-b\right)^2\ge0\forall a,b\)

Đẳng thức xảy ra khi \(\dfrac{1}{4}\left(2a-b\right)^2=0\Leftrightarrow2a=b\)