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 \(a_{12}\).

Answer 1

Recall Cauchy–Schwarz inequality :
\[ \langle x,x\rangle\cdot \langle y,y \rangle \ge |\langle x,y\rangle|^2\]

Let \(x=(1,\sqrt{2},\ldots,\sqrt{20})\)
\(y=(1a_1,\sqrt{2}a_2,\ldots,\sqrt{20}a_{20})\)

\((1+2+\cdots +20)\cdot (1a_1^2+2a_2^2+\cdots+20a_{20}^2)\ge(1a_1+2a_2+\cdots + 20a_{20})^2=1 \)
\(\implies 1a_1^2+2a_2^2+\cdots+20a_{20}^2\ge \dfrac{2}{20*21} \)

Answer 2

The equality is achieved when \(x\) and \(y\) are parallel :
\(y=kx\implies a_1=a_2=\cdots=a_{20}=a\)
\(\implies a(1+2+\cdots +20)=1 \implies a=\dfrac{2}{20*21}\)

so \(a_{12}=\dfrac{1}{210}\)