Maximizing the sum of two numbers, the sum of whose squares is constant 
How could we prove that if the sum of the squares of two numbers is a constant, then the sum of the numbers would have its maximum value  when the numbers are equal?

This result is also true for more than two numbers. I tested the results by taking various values and brute-force-ish approach. However I am interested to know a formal proof of the same. 
This is probably an mild extention to this problem. I encountered this result while searching for a easy solution for the same.

Here's the pedestrian, Calculus I method:
Let $x$ and $y$ be the two numbers. The condition "the sum of the squares is a constant" means that there is a fixed number $c$ such that $x^2+y^2=c$. That means that if you know one of the two numbers, say, $x$, then you can figure out the absolute value of the other one: $y^2 = c-x^2$, so $|y|=\sqrt{c-x^2}$. 
Now, since you want to find the maximum of the sum, then you can restrict yourself to the positive values of $x$ and $y$ (the condition on the sum of squares doesn't restrict the sign). So we may assume $x\geq 0$, $y\geq 0$, and $y=\sqrt{c-x^2}$.  And now you want to find the maximum of $x+y = x+\sqrt{c-x^2}$.
Thus, this reduces to finding the maximum value of $S(x) = x+\sqrt{c-x^2}$ on the interval $[0,\sqrt{c}]$. 
By the Extreme Value Theorem, the maximum will be achieved either at an endpoint or at a critical point of $S(x)$. At $0$ we get $S(0) = \sqrt{c}$; at $\sqrt{c}$ we get $S(\sqrt{c}) = \sqrt{c}$. As for critical points, 
$$S'(x) = 1 - \frac{x}{\sqrt{c-x^2}}.$$
The critical points are $x=\sqrt{c}$ (where $S'(x)$ is not defined), and the point where $x=\sqrt{c-x^2}$, or $2x^2=c$; hence $x^2=\frac{c}{2}$ (which means $y^2=\frac{c}{2}$ as well). At $x=\sqrt{\frac{c}{2}}$, $S(x) = 2\sqrt{\frac{c}{2}} = \sqrt{2c}$. This is clearly the maximum.
For the problem with $k$ variables, $k\gt 2$, the "pedestrian method" involves Multivariable calculus and Lagrange multipliers.
