Putnam 2000, Problem A-1

Let $A$ be a positive real number. What are the possible values of $\sum_{j=0}^{\infty} x_j^2$, given that $x_0, x_1, \ldots$ are positive numbers for which $\sum_{j=0}^{\infty} x_j = A$?

Solution:
The possible values comprise the interval \((0,A^2)\).  
To see that the values must lie in this interval, note that  
\[
\left( \sum_{j=0}^m x_j \right)^2 = \sum_{j=0}^m x_j^2 + \sum_{0 \leq j < k \leq m} 2x_jx_k,
\]  
so \(\sum_{j=0}^m x_j^2 \leq A^2 - 2x_0x_1\). Letting \(m \to \infty\), we have \(\sum_{j=0}^\infty x_j^2 \leq A^2 - 2x_0x_1 < A^2\).  

To show that all values in \((0,A^2)\) can be obtained, we use geometric progressions with \(x_1/x_0 = x_2/x_1 = \cdots = d\) for variable \(d\). Then \(\sum_{j=0}^\infty x_j = x_0/(1-d)\) and  
\[
\sum_{j=0}^\infty x_j^2 = \frac{x_0^2}{1-d^2} = \frac{1-d}{1+d} \left( \sum_{j=0}^\infty x_j \right)^2.
\]  

As \(d\) increases from \(0\) to \(1\), \((1-d)/(1+d)\) decreases from \(1\) to \(0\). Thus if we take geometric progressions with \(\sum_{j=0}^\infty x_j = A\), \(\sum_{j=0}^\infty x_j^2\) ranges from \(0\) to \(A^2\). Thus the possible values are indeed those in the interval \((0,A^2)\), as claimed.