Sửa đề: \(\sqrt{x-1}+\sqrt{5-x}+...\)
Đặt \(\sqrt{x-1}+\sqrt{5-x}=t\Rightarrow2\le t\le2\sqrt{2}\)
\(t^2=4+2\sqrt{-x^2+6x-5}\Rightarrow\sqrt{-x^2+6x-5}=\frac{1}{2}t^2-2\)
BPT trở thành: tìm m lớn nhất để
\(t+\frac{1}{2}t^2-2\ge m\) với mọi \(t\in\left[2;2\sqrt{2}\right]\)
Xét \(f\left(t\right)=\frac{1}{2}t^2+t-2\) trên \(\left[2;2\sqrt{2}\right]\)
\(-\frac{b}{2a}=-1\notin\left[2;2\sqrt{2}\right]\)
\(f\left(2\right)=-\frac{11}{8};f\left(2\sqrt{2}\right)=2+2\sqrt{2}\)
\(\Rightarrow\min\limits_{\left[2;2\sqrt{2}\right]}f\left(t\right)=f\left(2\right)=-\frac{11}{8}\)
\(\Rightarrow\) Để \(f\left(t\right)\ge m;\forall t\in\left[2;2\sqrt{2}\right]\Leftrightarrow m\le\min\limits_{\left[2;2\sqrt{2}\right]}f\left(t\right)=-\frac{11}{8}\)
\(\Rightarrow m_{max}=-\frac{11}{8}\)
