Assume that $ 1a_1+2a_2+\cdots+20a_{20}=1, $ where the $a_j$ are real numbers and that these values minimize...

Question

Assume that $\ 1a_1+2a_2+\cdots+20a_{20}=1, $ where the $a_j$ are real numbers and that these values minimize

$[1a_1^2+2a_2^2+\cdots+20a_{20}^2.]$ Find $a_{12}$.

Any help would be greatly appreciated. Thanks!

Answer 1

Use Cauchy-Schwarz on the vectors $$\mathbf{x} = (1, \sqrt{2}, \sqrt{3}, \dots, \sqrt{20})$$ and $$\mathbf{y} = (1a_1, \sqrt{2}a_2, \sqrt{3}a_3, \dots, \sqrt{20}a_{20})$$

So, $\mathbf{x}\cdot\mathbf{x} = 1 + 2 + 3 + \dots + 20 = 210$ and $\mathbf{y}\cdot\mathbf{y} = a_1^2 + 2a_2^2 + 3a_3^2 + \dots + 20 a_{20}^2$.

Also, $\mathbf{x}\cdot\mathbf{y} = a_1 + 2 a_2 + 3a_3 + \dots + 20 a_{20}=1$. The Cauchy-Schwarz inequality says $$(\mathbf{x}\cdot\mathbf{x})(\mathbf{y}\cdot\mathbf{y}) \geq (\mathbf{x}\cdot\mathbf{y})^2$$ with equality if and only if $\mathbf{x}$ and $\mathbf{y}$ are parallel. Hence, $$210(a_1^2 + 2a_2^2 + 3a_3^2 + \dots + 20 a_{20}^2) \geq 1$$

In particular, $a_1^2 + 2a_2^2 + 3a_3^2 + \dots + 20 a_{20}^2 \geq \frac{1}{210}$, so this is the minimum value. To have equality, we need $k \mathbf{x} = \mathbf{y}$ for some k, that is, $k \sqrt{i} = \sqrt{i} a_{i}$, hence $a_i = k$ for all $i$.

It finally remains to find the value of k, so use that $a_1 + 2 a_2 + 3a_3 + \dots + 20 a_{20}=1$, so $1 = k + 2 k + 3k + \dots + 20 k = 210k$, so $k = \frac{1}{210}$. In particular, $a_{12} = \frac{1}{210}$.