\(2\left(a^2+b^2\right)\ge a^2+b^2+2ab=\left(a+b\right)^2\left(\text{Cô si}\right)\)
\(\Rightarrow a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\) (Dấu "=" xảy ra⇔a=b)
Áp dụng, ta có:
\(F=\sqrt{\left(x-2011\right)^2+\left(x-1\right)^2}\)
\(=\sqrt{\left(2011-x\right)^2+\left(x-1\right)^2}\)
\(\ge\sqrt{\dfrac{1}{2}\left(2011-x+x-1\right)^2}=\sqrt{\dfrac{1}{2}\left(2010\right)^2}=\sqrt{\dfrac{2010^2}{2}}=\sqrt{\dfrac{1005^2.2^2}{2}}=1005\sqrt{2}\)
Dấu"=" xảy ra\(\Leftrightarrow x-2011=x-1\)
\(\Leftrightarrow0=2010\left(\text{vô lí}\right)\)
⇒ không xảy ra dấu "="
Sửa chỗ dấu bằng nha, a lm nhầm
\(\Leftrightarrow2011-x=x-1\)
\(\Leftrightarrow x=1006\)
Vậy Min\(F=1005\sqrt{2}\Leftrightarrow x=1006\)
