## zmudz one year ago Assume that $$1a_1+2a_2+\cdots+na_n=1,$$ where the $$a_j$$ are real numbers. As a function of $$n$$, what is the minimum value of $$1a_1^2+2a_2^2+\cdots+na_n^2?$$ I tried 1/5 and then 5, but none of them work. Please help! Thanks!

I'm pretty sure you can use Lagrange multipliers to solve this. Let $$S_1=\sum\limits_{j=1}^nja_j$$ and $$S_2=\sum\limits_{j=1}^nj{a_j}^2$$. You're looking to minimize $$S_2$$ with respect to the constraint $$S_1$$. The Lagrangian is given by $\mathcal{L}(a_1,\ldots,a_n,\lambda)=S_2+\lambda S_1$ Take the gradient and set equal to $$0$$: \left.\begin{align*} \frac{\partial}{\partial a_1}\left[S_2+\lambda S_1\right]&=2a_1+\lambda\\[2ex] \frac{\partial}{\partial a_2}\left[S_2+\lambda S_1\right]&=4a_2+2\lambda\\[2ex] &\vdots\\ \frac{\partial}{\partial a_n}\left[S_2+\lambda S_1\right]&=2na_n+n\lambda\\[2ex] \frac{\partial}{\partial \lambda}\left[S_2+\lambda S_1\right]&=S_1 \end{align*}\right\}=0\quad(\text{each}) The first $$n$$ equations tell you that $$a_j=-\dfrac{\lambda}{2}$$. Plugging this into the original constraint, you get $S_1=\sum_{j=1}^n ja_j=-\frac{\lambda}{2}\sum_{j=1}^n j=-\frac{\lambda n(n+1)}{4}=1$or $$\lambda=-\dfrac{4}{n(n+1)}$$. This means $$S_2$$ is minimized when $a_j=-\frac{\lambda}{2}=\frac{2}{n(n+1)}$ Compute the sum: \begin{align*}\sum_{j=1}^n j{a_j}^2&=\sum_{j=1}^nj\left(\frac{2}{n(n+1)}\right)^2\\[1ex] &=\frac{4}{n^2(n+1)^2}\sum_{j=1}^nj\\[1ex] &=\frac{4}{n^2(n+1)^2}\cdot\frac{n(n+1)}{2}\\[1ex] &=\frac{2}{n(n+1)}\end{align*}