The minimum value of f(x)=(x-a)^2+(x-b)^2+(x-c)^2 is

The correct answer is Option (1) → \( \displaystyle\frac{1}{9}\left[(b+c-2a)^2+(c+a-2b)^2+(a+b-2c)^2\right] \)

The minimum comes from the first derivative test, confirmed by the second derivative.

    \( \displaystyle f'(x)=2(x-a)+2(x-b)+2(x-c) \)

    \( \displaystyle =2\left[3x-(a+b+c)\right] \)

    \( \displaystyle f'(x)=0\ \Rightarrow\ x=\frac{a+b+c}{3} \)

    \( \displaystyle f''(x)=6\gt 0 \)

So f has its minimum at \( x=\frac{a+b+c}{3} \). At this point

    \( \displaystyle x-a=\frac{a+b+c}{3}-a=\frac{b+c-2a}{3} \)

and in the same way

    \( \displaystyle x-b=\frac{c+a-2b}{3} \)

    \( \displaystyle x-c=\frac{a+b-2c}{3} \)

Squaring and adding, each square brings a factor \( \frac{1}{9} \):

    \( \displaystyle f_{\min}=\frac{1}{9}\left[(b+c-2a)^2+(c+a-2b)^2+(a+b-2c)^2\right] \)

Tip: A sum of squared distances \( \sum(x-a_i)^2 \) is always smallest when x is the mean of the \( a_i \). Remembering this saves the differentiation.