\(\sqrt{7-x}+\sqrt{x+1}=x^2-6x+13\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(VT^2=\left(\sqrt{7-x}+\sqrt{x+1}\right)^2\)
\(\le\left(1+1\right)\left(7-x+x+1\right)=16\)
\(\Rightarrow VT^2\le16\Rightarrow VT\le4\)
Lại có: \(VP=x^2-6x+13\)
\(=x^2-6x+9+4=\left(x-3\right)^2+4\ge4\)
Suy ra \(VT\le VP=4\) xảy ra khi \(VT=VP=4\)
\(\Rightarrow\left(x-3\right)^2+4=4\Rightarrow x-3=0\Rightarrow x=3\)