$a_1 + a_2 + \dots a_n = 1$ find min of $a_1^2 +\frac{a_2^2}{2} + \dots + \frac{a_n^2}n.$

Question

Given $n$ numbers $a_1$, $\cdots$ and such that $a_1$, $a_2$, $\cdots$, $a_n > 0$ and their sum is $1$, I want to find the minimum value of

$$a_1^2 + \frac{a_2^2}{2} + \cdots + \frac{a_n^2}{n}.$$

I have tried using weighted AM-GM inequality, like this:

$$\frac{a_1^2 + \frac{a_2^2}{2} + \cdots + \frac{a_n^2}{n}}{a_1 + a_2 + \cdots + a_n } \geqslant \frac{a_1^{a_1} \cdots a_n^{a_n}}{2^{a_2} \cdots n^{a_n}}$$

but was unable to make progress on the right hand side.

Is there a better way to apply AM-GM inequality? Or is there some different way altogether to solve this?

Answer 1

By the Cauchy-Schwarz inequality we have $$ \left(\sum_{i=1}^{n}i\right)\left(\sum_{i=1}^{n}\frac{a_i^2}{i}\right) \geq \left(\sum_{i=1}^{n}\sqrt{i}\cdot\frac{a_i}{\sqrt{i}}\right)^2 = 1 \tag{1}$$ hence with our hypothesis we have: $$ \sum_{i=1}^{n}\frac{a_i^2}{i}\geq \frac{2}{n(n+1)}\tag{2} $$ and equality is achieved iff $\frac{a_i}{\sqrt{i}}=\lambda\sqrt{i}$, i.e. iff $a_i = \frac{2i}{n(n+1)}$.

Answer 2

Hint: Apply the Cauchy-Schwarz Inequality to $$\left(\sum_{i=1}^n\,i\right)\,\left(\sum_{i=1}^n\,\frac{a_i^2}{i}\right)\,.$$ Alternatively, apply the Power-Mean Inequality to $$\sqrt{\frac{\sum_{i=1}^n\,i\,\left(\frac{a_i}{i}\right)^2}{\sum_{i=1}^n\,i}}\,.$$

Answer 3

Consider the continuous function $f(a_1,a_2\dots,a_n)=a_1^2+a_2^2/2+\dots +a_n^2/n$. We will minimize it in the compact region defined by constraints $a_i\geq 0$ and $a_1+a_2+\dots + a_n=1$.

The idea is that since the region is compact the minimum must exist, and we will find necessary conditions for the point at which the minimum is achieved.

Suppose that $a_1+a_k=z$, then $a_1^2+\frac{a_k^2}{k}$ must be maximized with respect to $a_1+a_k=z$, when does this happen? we want to maximize $a_1^2+\frac{(z-a_1)^2}{k}$,the derivative of this function is $2a_1-\frac{2(z-a_1)}{k}$. So the minimum is reached when $(k+1)a_1=z$ or $a_1=\frac{z}{k+1}$, or when $ka_1=a_k$.

We conclude the maximum is reached at:

$a_1,2a_1,3a_1\dots na_1$, which clearly adds up to $1$ when $a_1=\frac{2}{n(n+1)}$.

Answer 4

I would go with Cauchy Schwarz inequality $$\left(\sum_{cyc} \frac{a_i^2}i\right)(1+2+\dots + n) \geqslant \sum_{cyc} a_i = 1$$

Gives the minimum to be $\dfrac2{n(n+1)}$ directly, as equality is possible when $a_i$ is proportional to $i$.

In case you want to use AM-GM, you may sum $n$ equations of kind $$\frac{a_i^2}i + \frac{4i}{n^2(n+1)^2} \geqslant \frac{4a_i}{n(n+1)} \implies \sum \frac{a_i^2}i \geqslant \frac4{n(n+1)} - \frac2{n(n+1)} = \frac2{n(n+1)}$$

Answer 5

we have $$a_1^2+\frac{a_2^2}{2}+...\frac{a_n^2}{n}\geq \frac{(a_1+a_2+...+a_n)^2}{1+2+...+n}\geq \frac{1}{1+2+...+n}=\frac{1}{\frac{n(n+1)}{2}}=\frac{2}{n(n+1)}$$