\(T=\sqrt{x^2-x+2}+\sqrt{x^2+x+2}\)
\(T^2=x^2-x+2+x^2+x+2+2\sqrt{\left(x^2-x+2\right)\left(x^2+x+2\right)}\)
\(T^2=2x^2+4+2\sqrt{\left(x^2+2\right)^2-x^2}\)
\(T^2=2x^2+4+2\sqrt{x^4+4x^2+4-x^2}\)
\(T^2=2x^2+4+2\sqrt{x^4+3x^2+4}\)
Nhận xét: \(2x^2\ge0\forall x\)
\(x^4+3x^2+4=x^2\left(x^2+3\right)+4\)
Có: \(x^2\ge0,x^2+3\ge0\forall x\) nên:
\(\Rightarrow x^2\left(x^2+3\right)\ge0\forall x\)
Cho nên: \(x^4+3x^2+4\ge4\)
Vậy \(T^2=2x^2+4+2\sqrt{x^4+3x^2+4}\ge4+2\sqrt{4}=4+4=8\)
Do \(T^2\ge8\) nên:
\(\Rightarrow A\ge2\sqrt{2}\)
Dấu \("="\) xảy ra \(\Leftrightarrow x=0\)
Cái này dùng Min-cốp-xki được nè
\(T=\sqrt{\left(\frac{1}{2}-x\right)^2+\left(\frac{\sqrt{7}}{2}\right)^2}+\sqrt{\left(\frac{1}{2}+x\right)+\left(\frac{\sqrt{7}}{2}\right)^2}\ge\sqrt{\left(\frac{1}{2}-x+\frac{1}{2}+x\right)+\left(\sqrt{7}\right)^2}=\sqrt{1^2+7}=2\sqrt{2}\)
Dấu "=" xảy ra <=> \(\frac{1}{2}-x=\frac{1}{2}+x\)
<=> x = 0
