What is the minimum value of $\sum_{n=1}^{\infty}a_n^2$ $\forall a_n >0$ such that $\sum_{n=1}^{\infty}a_n=A$?

Question

I don't think Lagrangian multiplier method can be used with infinite sums. I suspect this could be solved via cauchy-schwarz inequality, but I am not able to think through.

Answer 1

By generalized mean $M_2\ge M_1$ we have that

$$\sqrt{\frac{\sum a_i^2}{n}}\ge \frac{\sum a_i}{n}=\frac A n \iff \sum a_i^2\ge \frac{A^2}n$$

and equality holds for $a_1=\ldots=a_n=\frac A n$.

Therefore in the limit the minimum value tends to $0$ but since

$$\sum_{i=1}^{\infty}a_i^2 =0 \implies \forall i \quad a_i =0 $$

according to the given condition the value is not reached and we have

$$\sum_{i=1}^{\infty}a_i^2 >0$$

Answer 2

Starting from achille hui's comment, for $N\geq 2$ and $n\geq 1$, take $$a_n:=\frac{A}{N}\cdot \begin{cases} 1-\frac{1}{N}&\text{if $n=1,\dots,N,$}\\ \frac{1}{2^{n-N}}&\text{if $n\geq N+1.$} \end{cases}$$ Then it is easy to verify that $a_n>0$ for all $n\geq 1$, $\sum_{n=1}^{\infty}a_n=A$ and $$\sum_{n=1}^{\infty}a_n^2=\frac{A^2}{N^2}\left(N\left(1-\frac{1}{N}\right)^2+\frac{1}{3}\right)$$ which goes to zero as $N$ goes to infinity. Hence the desired infimum is zero (which is not attained since $\sum_{n=1}^{\infty}a_n^2=0$ implies that $a_n=0$ for all $n\geq 1$).