\(\sqrt{x^2-4x+5}+\sqrt{x^2-4x+8}+\sqrt{x^2-4x+9}=3+\sqrt{5}\)
\(\Leftrightarrow\sqrt{\left(x-2\right)^2+1}+\sqrt{\left(x-2\right)^2+4}+\sqrt{\left(x-2\right)^2+5}=3+\sqrt{5}\)
Ta có : \(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\forall x\\\left(x-2\right)^2\ge0\forall x\\\left(x-2\right)^2\ge0\forall x\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^2+1\ge1\\\left(x-2\right)^2+4\ge4\\\left(x-2\right)^2+5\ge5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{\left(x+2\right)^2+1}\ge1\\\sqrt{\left(x+2\right)^2+4}\ge2\\\sqrt{\left(x-2\right)^2+5}\ge\sqrt{5}\end{matrix}\right.\)\(\Rightarrow\sqrt{\left(x-2\right)^2+1}+\sqrt{\left(x-2\right)^2+4}+\sqrt{\left(x-2\right)^2+5}\ge3+\sqrt{5}\)
Đẳng thức xảy ra khi : \(x-2=0\Leftrightarrow x=2\)
KL..........
