Trên \(\left[0;3\right]\) do \(x^3-2x^2+2>0\) nên:
\(\Leftrightarrow ln\left(x^3-2x^2+2\right)-ln\left(x^2+m\right)+\left(x^3-2x^2+2\right)-\left(x^2+m\right)\ge0\)
\(\Leftrightarrow ln\left(x^3-2x^2+2\right)+\left(x^3-2x^2+2\right)\ge ln\left(x^2+m\right)+\left(x^2+m\right)\) (1)
Xét hàm \(f\left(t\right)=lnt+t\) với \(t>0\)
\(f'\left(t\right)=\dfrac{1}{t}+1>0\Rightarrow f\left(t\right)\) đồng biến
Do đó (1) tương đương \(x^3-2x^2+2\ge x^2+m\)
\(\Leftrightarrow m\le x^3-3x^2+2\)
\(\Leftrightarrow m\le\min\limits_{\left[0;3\right]}f\left(x\right)\) với \(f\left(x\right)=x^3-3x^2+2\)
\(f'\left(x\right)=3x^2-4x=0\Rightarrow x=\left\{0;\dfrac{4}{3}\right\}\)
\(f\left(0\right)=2\) ; \(f\left(\dfrac{4}{3}\right)=-\dfrac{26}{27}\); \(f\left(3\right)=2\)
\(\Rightarrow\min\limits_{\left[0;3\right]}f\left(x\right)=-\dfrac{26}{27}\Rightarrow m\le-\dfrac{26}{27}\)
