\(\Leftrightarrow\sqrt{\left(x-2\right)^2+1}\sqrt{\left(x-2\right)^2+4}+\sqrt{\left(x-2\right)^2+5}=3+\sqrt{5}\)
Có: \(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\forall x\\\left(x-2\right)^2\ge0\forall x\\\left(x-2\right)^2\ge0\forall x\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^2+1\ge1\\\left(x-2\right)^2+4\ge4\\\left(x-2\right)^2+5\ge5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{\left(x+2\right)^2+1}\ge1\\\sqrt{\left(x+2\right)^2+4\ge2}\\\sqrt{\left(x-2\right)^2+5}\ge\sqrt{5}\end{matrix}\right.\)
\(\Rightarrow\sqrt{\left(x+2\right)^2+1}\ge1+\sqrt{\left(x+2\right)^2+4}\ge2+\sqrt{\left(x-2\right)^2+5}\ge\sqrt{5}=3+\sqrt{5}\)Đẳng thức xảy ra khi: \(x-2=0\Leftrightarrow x=2\)
Vậy...