Ta có: \(\left(x-1\right)^3\ge\dfrac{3}{4}x-1\)
\(\Leftrightarrow x^3-3x^2+3x-1+1-\dfrac{3}{4}x\ge0\)
\(\Leftrightarrow x\left(x^2-3x+\dfrac{9}{4}\right)\ge0\)
\(\Leftrightarrow x\left[x\left(x-\dfrac{3}{2}\right)-\dfrac{3}{2}\left(x-\dfrac{3}{2}\right)\right]\ge0\)
\(\Leftrightarrow x\left(x-\dfrac{3}{2}\right)^2\ge0\) (Luôn đúng vì \(x\ge0\) và \(\left(x-\dfrac{3}{2}\right)^2\ge0\)
\(\Rightarrowđpcm\)
Đúng 0
Bình luận (2)
