Generalizing stuff

View each pair $(a_i,b_i)$ as a ordered pair in the Cartesian plane. Now each square root $\sqrt{a_i^2+b_i^2}$ represents the distance from the origin to that point (the hypotenuse).

Let $p = \sum a_i$ and $q = \sum b_i.$ Assume that we put each hypotenuse tip to tail so that they extend in a line. This "trail of hypotenuses" no matter what each individual length is, will end at the point $(p,q)$

Thus the overall length this trial will represent our sum. Which is obviously minimized when this length is straight, so the minimum value is $p^2+q^2.$

Using the Power Mean (QAGH) inequalities, $Q.M.\geq A.M.$ :

$\sqrt{\dfrac{p^2+q^2}{2}}\geq \dfrac{p+q}{2}$

Since $p+q=10,$

$p^2+q^2\geq \boxed{50}$

The equality can be achieved when $a_1=a_2=\cdots = a_n = b_1=b_2 =\cdots = b_n = \frac{10}{2n}.$

A solution using Lagrange multipliers seems to work here. Since the objective function is positive, it's not necessary to deal with the square, so just ignore it and minimize $\sum_{i=1}^n \sqrt{a_i^2+b_i^2}$ . You end up with the two sets of equations: $\frac{a_i}{\sqrt{a_i^2+b_i^2}}=\lambda$ and $\frac{b_i}{\sqrt{a_i^2+b_i^2}}=\lambda$ . But from here, it can be seen that $a_i=b_i$ . It doesn't matter what values you choose for $a_i,b_i$ as long as they add up to $10$ and $a_i=b_i$ (zero seems to be permissible as well); you'll end up with the sought minimum value. For example, $a_1=b_1=2,a_2=b_2=3$ , with the rest zero, has a sum of $10$ and gives $(\sqrt{2^2+2^2}+\sqrt{3^2+3^2})^2=50$ .