\(\sqrt{x^2-4x+5}+\sqrt{x^2-4x+8}+\sqrt{x^2-4x+9}=3+\sqrt{5}\\ \Leftrightarrow\sqrt{\left(x^2-4x+4\right)+1}+\sqrt{\left(x^2-4x+4\right)+4}+\sqrt{\left(x^2-4x+4\right)+5}=3+\sqrt{5}\\ \Leftrightarrow\sqrt{\left(x-2\right)^2+1}+\sqrt{\left(x-2\right)^2+4}+\sqrt{\left(x-2\right)^2+5}=3+\sqrt{5}\)Do \(\left(x-2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-2\right)^2+1\ge1\forall x\\ \left(x-2\right)^2+4\ge4\forall x\\ \left(x-2\right)^2+5\ge5\forall x\\ \Rightarrow\sqrt{\left(x-2\right)^2+1}\ge1\forall x\\ \sqrt{\left(x-2\right)^2+4}\ge2\forall x\\ \sqrt{\left(x-2\right)^2+5}\ge\sqrt{5}\forall x\)
\(\Rightarrow\sqrt{\left(x-2\right)^2+1}+\sqrt{\left(x-2\right)^2+4}+\sqrt{\left(x-2\right)^2+5}\ge1+2+\sqrt{5}\ge3+\sqrt{5}\)
Dấu "=" xảy ra khi:
\(x-2=0\\ \Leftrightarrow x=2\)
Vậy...........
