Minimum possible sum of squares of two numbers with sum $k$? If the sum of two numbers is k. Find the minimum value of the sum of their squares.
This is my calculations so far.
a + b = k                <-- google says that I should put x + y = k rather than a + b = k
a² + b² = y                  (but I don't know why should I do that)
(k - b)² + b² = y
k² - 2kb + 2b² = y       <-- now I'm stuck here. I don't know which differentiation variable to take. If it is 'b' then why?

So here I am. Kindly show me how to solve the problem and where did I go wrong in my computations :P

Very simple approach:
You know $a + b = k$.  So $$(a+b)^2 = a^2 + 2ab + b^2 = k^2.$$  If you could somehow remove the $2ab$ term, then you'd have the desired sum of squares.  Well, you also know that $$(a-b)^2 = a^2 - 2ab + b^2,$$ so if you add the two together, you'd get
$$(a+b)^2 + (a-b)^2 = 2a^2 + 2b^2 = k^2 + (a-b)^2.$$
Therefore, $$a^2 + b^2 = \frac{k^2 + (a-b)^2}{2}.$$
Now, $k$ is a constant, and $(a-b)^2$, being the square of a real number, is never negative, so the right hand side is minimized when you can make $a-b = 0$, and the minimum value attained is $k^2/2$.
