Question

Với 0 < x < \(\dfrac{4}{3}\)

Chứng minh rằng : \(\dfrac{1}{x^2\left(4-3x\right)}\ge x\)

Answer 1

\(BDT\Leftrightarrow\dfrac{1}{x^2\left(4-3x\right)}-x\ge0\)

\(\Leftrightarrow\dfrac{1}{x^2\left(4-3x\right)}-\dfrac{x^3\left(4-3x\right)}{x^2\left(4-3x\right)}\ge0\)

\(\Leftrightarrow\dfrac{1-3x^4+4x^3}{x^2\left(4-3x\right)}\ge0\)

\(\Leftrightarrow\dfrac{\left(x-1\right)^2\left(3x^2+2x+1\right)}{x^2\left(4-3x\right)}\ge0\)

\(\Leftrightarrow\dfrac{\left(x-1\right)^2\left(3\left(x+\dfrac{1}{3}\right)^2+\dfrac{2}{3}\right)}{x^2\left(4-3x\right)}\ge0\forall0< x< \dfrac{4}{3}\)