Does $\sum\limits_{i=1}^n x_i = 1$ imply $\sum\limits_{i=1}^n x_i^2 \geq \frac{1}{n}$?

Question

Suppose we have real numbers $x_1, ..., x_n$ which satisfy $x_1 + ... + x_n = 1$. Do we have the lower bound $x_1^2 + ... + x_n^2 \geq \frac{1}{n}$?

It seems intuitive that we can minimize this by setting all the $x_i$ to be $\frac{1}{n}$, but I can't seem to prove it.

I also couldn't find a reference to this result. I would appreciate it if anyone could provide one!

Answer 1

You can prove that $$x_i^2\ge \frac 2n x_i - \frac{1}{n^2}$$ and sum up for $i=1,2,\ldots,n$.

Answer 2

Use that $$\frac{x_1+x_2+...+x_n}{n}\le \sqrt{\frac{x_1^2+x_2^2+...+x_n^2}{n}}$$ .

Answer 3

By C-S $$\sum_{k=1}^nx_k^2=\frac{1}{n}\sum_{k=1}^n1\sum_{k=1}^nx_k^2\geq\frac{1}{n}\left(\sum_{k=1}^nx_k\right)^2=\frac{1}{n}.$$ We can use also the Tangent Line method: $$\sum_{k=1}^nx_k^2-\frac{1}{n}=\sum_{k=1}^n\left(x_k^2-\frac{1}{n^2}\right)=\sum_{k=1}^n\left(x_k-\frac{1}{n}\right)\left(x_k+\frac{1}{n}\right)=$$ $$=\sum_{k=1}^n\left(\left(x_k-\frac{1}{n}\right)\left(x_k+\frac{1}{n}\right)-\frac{2}{n}\left(x_k-\frac{1}{n}\right)\right)=\sum_{k=1}^n\left(x_k-\frac{1}{n}\right)^2\geq0.$$

Answer 4

Basically use the variance is non-negative bound. That is for any random variable, $$E[X^2]\ge E[X]^2$$

Let $X$ be a random variable taking values $x_1$ to $x_n$ with equal probability. Can you handle the rest?

Answer 5

Another useful inequality is Chebyshev's one $$\sum\limits_{i=1}^n a_ib_i\geq \frac{1}{n}\left(\sum\limits_{i=1}^n a_i\right)\left(\sum\limits_{i=1}^n b_i\right)$$ for $a_1\geq a_2 \geq ... \geq a_n$ and $b_1\geq b_2 \geq ... \geq b_n$

We can always re-organise $x_i$ in descending order $x_1\geq x_2 \geq ... \geq x_n$ and this won't affect $\sum\limits_{i=1}^n x_i$ and $\sum\limits_{i=1}^n x_i^2$ values. As a result

$$\sum\limits_{i=1}^n x_i^2= \sum\limits_{i=1}^n x_ix_i \geq \frac{1}{n}\left(\sum\limits_{i=1}^n x_i\right)\left(\sum\limits_{i=1}^n x_i\right)=\frac{1}{n}$$