Upper bound to sum of squares using given sum

Question

If $\sum_{i = 1}^n a_i = x$ for $a_i \geq 0$, then is it possible to find upper bound to $\sum_{i=1}^n a_i^2$? I know that the lower bound can be easily determined using Cauchy- Schwarz Inequality. But how to derive the upper bound?

Answer 1

Do you assume that $a_i\ge 0$ ?

Otherwise take $a_1+a_2=0$ with $a_1=t$, $a_2=-t$, then $a_1^2+a_2^2=2t^2$ has no upper bound.

Answer 2

One simple bound is $$\sum_{i=1}^{n} a_i^2 \leq \sum_{i=1}^{n}\sum_{j=1}^{n}a_ia_j = \left(\sum_{i=1}^{n} a_i\right) \left(\sum_{j=1}^{n} a_j\right) = x^2.$$ The inequality holds because the LHS is the sum of just the $i=j$ terms in the double sum.

This bound is achieved if and only if $a_i a_j = 0$ whenever $i \neq j$, which is true if and only if at most one of the $a_i$'s is nonzero.