ĐK -1 =< x =< 3
\(T^2=x+1+2\sqrt{\left(x+1\right)\left(3-x\right)}+3-x=4+2\sqrt{3x-x^2+3-x}\)
\(=4+2\sqrt{-x^2+2x+3}=4+2\sqrt{-\left(x^2-2x+1-1\right)+3}\)
\(=4+2\sqrt{-\left(x-1\right)^2+4}\)
Ta có \(-\left(x-1\right)^2+4\le4\Leftrightarrow2\sqrt{-\left(x-1\right)^2+4}\le-2\Leftrightarrow4+2\sqrt{-\left(x-1\right)^2+4}\le2\)
=> T =< \(\sqrt{2}\)Dấu ''='' xảy ra khi x = 1
Ta có BĐT: \(\sqrt{A}+\sqrt{B}\ge\sqrt{A+B}\left(A,B\ge0\right)\)
Dấu "=" xảy ra \(\Leftrightarrow\left[{}\begin{matrix}A=0\\B=0\end{matrix}\right.\)
- Áp dụng:
\(ĐKXĐ:-1\le x\le3\)
\(T=\sqrt{x+1}+\sqrt{3-x}\ge\sqrt{x+1+3-x}=\sqrt{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
Vậy \(MinT=\sqrt{2}\)
