\(\Leftrightarrow-x^2+2x+8-4\sqrt{-x^2+2x+8}+10\le m\)
Đặt \(\sqrt{-x^2+2x+8}=t\Rightarrow t\in\left[0;8\right]\)
\(\Rightarrow t^2-4t+10\le m\)
Xét \(f\left(t\right)=t^2-4t+10\) trên \(\left[0;8\right]\)
\(a=1>0\) ; \(-\frac{b}{2a}=2\in\left[0;8\right]\)
\(f\left(0\right)=10\) ; \(f\left(2\right)=6\) ; \(f\left(8\right)=42\)
\(\Rightarrow6\le f\left(t\right)\le42\)
Để \(f\left(t\right)\le m;\forall t\in\left[0;8\right]\Leftrightarrow m\ge\max\limits_{\left[0;8\right]}f\left(t\right)\)
\(\Rightarrow m\ge42\)