## zmudz one year ago 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 $$a_{12}$$.

1. ganeshie8

Recall Cauchy–Schwarz inequality : $\langle x,x\rangle\cdot \langle y,y \rangle \ge |\langle x,y\rangle|^2$ Let $$x=(1,\sqrt{2},\ldots,\sqrt{20})$$ $$y=(1a_1,\sqrt{2}a_2,\ldots,\sqrt{20}a_{20})$$ $$(1+2+\cdots +20)\cdot (1a_1^2+2a_2^2+\cdots+20a_{20}^2)\ge(1a_1+2a_2+\cdots + 20a_{20})^2=1$$ $$\implies 1a_1^2+2a_2^2+\cdots+20a_{20}^2\ge \dfrac{2}{20*21}$$

2. ganeshie8

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 $$a_{12}=\dfrac{1}{210}$$