The least possible value How to find the least possible value for :$$(x-1)^2+(x-2)^2+(x-3)^2+(x-4)^2+(x-5)^2$$
For every real $x$
 A: $(x-1)^2+(x-2)^2+(x-3)^2+(x-4)^2+(x-5)^2$
$=(y+2)^2+(y+1)^2+y^2+(y-1)^2+(y-2)^2$.Let $(y=(x-3))$
$=5y^2+2(1^2+2^2)\ge2(1^2+2^2)$
Equality occurs at $y=0\Rightarrow x=3$
A: Assuming you don't have any calculus knowledge, you can use techniques learned in algebra/pre-calculus.
You can square each term in $(x-1)^2+(x-2)^2+(x-3)^2+(x-4)^2+(x-5)^2$ and collect like terms to get the form $ax^2 + bx + c$.
Since $a$ will be positive, we know this is a parabola opening up, like:

The highest or lowest point on a parabola is the vertex. Since this parabola opens upward, we know that we will have a minimum at the vertex.
To find $x$, use $\displaystyle x = -\frac{b}{2a}$. This is the $x$ that gives the minimum value. Plug this back into your $ax^2 + bx + c$ expression to find the minimum value.
A: Hint: A twice differentiable function $f(x)$ is minimised iff $f'(x) = 0$ and $f''(x)>0$
Alternative hint: Expand out the brackets and write the expression in the form $ax^2+bx+c$ and then complete the square.
A: Hint:
$(x-1)^2+(x-2)^2+(x-3)^2+(x-4)^2+(x-5)^2 \ge 0$
$5x^2+1+4+9+16+25 \ge 15x$
$5x^2 +55 -15x \ge 0 \implies x^2+11-3x \ge 0$
Aliter: $(x-3)=k$
You get 
$(k+2)^2+(k+1)^2+(k)^2+(k-1)^2+(k-2)^2 \ge 0$
$k^2+11 \ge 0$
