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
 A: 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 $k^2/2$.
A: Hint: $x^2+y^2 \ge \dfrac{(x+y)^2}{2}$.
A: I focus on a calculus method given your tags on the question. We are given $a + b = k$. The sum of their squares is given by $a^2 + b^2$. Since $a + b = k$, we hat that $k - b = a$, hence
$$g(b) = a^2 + b^2 = (k-b)^2 + b^2 = k^2-2kb+b^2+b^2 = k^2 - 2kb + 2b^2\text{.}$$
$k$ is a fixed value that is known, so $b$ is the only unknown in the expression above. Differentiating the above expression with respect to $b$ and setting that equal to $0$, we obtain
$$g^{\prime}(b) = -2k + 4b = 0$$
or
$$4b = 2k \implies b = \dfrac{2}{4}k = \dfrac{k}{2} \text{.}$$
Since $b = \dfrac{k}{2}$ and $a + b = k$, it follows that $$a = k - b = k - \dfrac{k}{2} = \dfrac{k}{2}\text{.}$$
Hence from setting the derivative equal to $0$, we obtain $a = b = \dfrac{k}{2}$.
Next, we must demonstrate that this solution is a minimum. We have
$$g^{\prime\prime}(b) = 4 > 0$$
for any $b$, hence by the second derivative test, the solution $a = b = \dfrac{k}{2}$ must be a minimum.
Thus, the desired sum of squares is given by
$$a^2 + b^2 = \left(\dfrac{k}{2}\right)^2 + \left(\dfrac{k}{2}\right)^2 = 2 \cdot \dfrac{k^2}{4} = \dfrac{k^2}{2}\text{.}$$
A: Let $k = x + y$. This implies that $y = k - x$. The sum of their squares is $x^2 + y^2$, which can be written as $x^2 + (k - x)^2$. Since we are trying to to find the minimum value of this expression for all $x$, we must first determine for what $x$ values this expression's derivative (w.r.t. $x$) changes from negative to positive.
Set this expression equal to $f(x)$ and find $f'(x)$ and $f''(x)$:
$f(x) = x^2 + (k-x)^2$
$f'(x) = 2x - 2(k - x)$.
$f''(x) = 4$
To determine when $f'(x)$ changes from negative to positive on a quadratic (or any "non-linear" function), test for the condition $f'(x) = 0 \land f''(x) > 0$.
$f'(x) = 0 \equiv 0 = 2x - 2(k - x)$; $0 = 2x - 2k + 2x$; $2k = 4x$; $x = \frac{k}{2}$.
$f''(x) > 0 \equiv 4 > 0$, which is true regardless of the value of $x$.
Therefore, $f(x)$ has a relative minimum at $x = \frac{k}{2}$.
To find the global minimum, find the minimum value of the relative minimums of $f$ and $f$ evaluated at it's given bounds (if they are given). Since bounds are not given, test if $f(x)$ as $x \to \infty$ or $x \to - \infty$ are less than the candidate $\frac{k}{2}$. $f(x)$ is merely a sum of two squares, thus the limit must approach $\infty$. Therefore, as the only possible case, $\frac{k}{2}$ is the global minimum value of $x^2 + y^2$ given two numbers $x$, $y$ s.t. $x + y = k$.
