Áp dụng bđt bu-nhi-a cho VT ta có:
\(\left(\sqrt{x^2+x-1}\right)^2+\left(\sqrt{-x^2+x+1}\right)^2\ge\frac{\left(\sqrt{x^2+x-1}+\sqrt{-x^2+x+1}\right)^2}{2}\)
\(\Leftrightarrow\)\(x^2+x-1-x^2+x+1\ge\frac{VT^2}{2}\)
=>VT^2\(\le\)4x
=>VT\(\le\)\(2\sqrt{x}\)\(\le\)x+1
Lại có:VP=x^2-x+2\(\ge\)x+1
Mà VT=VP => VT=VT=x+1
Dấu "=" xảy ra <=>x=1