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Bunuel
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Maxima and minima

Used derivatives here

(x-a)^2+(x-b)^2

First derivative=> 2*(x-a)+2*(x-b) =0 =>x=(a+b)/2 ;substitute b=> x=a+2
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For a quadratic equation y = ax^2 + bx + c :

If a > 0 [Upward opening parabola]
y is min at (-b/2a)

If a < 0 [Downward opening parabola]
y is max at (-b/2a)

In this case upon getting the final equation after simplifying the squares and subtituting a = b+4,
We get -b/2a to be a+2
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One quick way to solve this problem is using the number line method.

Let a=2, b= a+4 = 6

We have to make (x−a)^2+(x−b)^2 minimum. Or in other words, the sum of the squared distance of x from a and the squared distance of x from b has to be smallest.

If you draw this on the number line, you'll quickly get that this value is smallest when x is 4, i.e, x= a+2.
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I made an attempt at using the inequality Arithmetic Mean is >= Geometric Mean. So then the least value for the expression is 2 * (x-a) * (x-b). Only a+2 and a+3 are contenders, a+2 wins as -6 is bigger than -8.
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Substituting the value of b will give :
(x-a)^2 + ((x-a)-4)^2

Now, opening the second term and simplifying:
2*(x-a)^2 -8(x-a) +16.

It will be easier now to check substituting the values of x as we have to necessarily check where the sum of first 2 terms is most negative which will be at x= a+2
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KarishmaB usually plugging values is not my first approach hence wasted time here - any other solution?
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