# For real $a_i$, does the inequality $\sum^N (a_i)^2 \geq \frac{1}{N}$ hold, if $\sum^N a_i = 1$

I saw this post here, stating

for real $x,y,z$ where $x + y + z = 1$, then $x^2 + y^2 + z^2 \geq \frac{1}{3}.$

So I was curious if we could generalize this further to any number of variables

Proposition For real $a_i$ where $\sum^N a_i = 1$, then $\sum^N (a_i)^2 \geq \frac{1}{N}$ for $N \in \mathbb{N}.$

The selected answer uses Cauchy-Schwarz inequality, which I'm not sure how to apply here. Another idea I got from that post was to expand $\sum^N (Na_i - 1)^2 \geq 0$ and simplify.

Expanded: $\sum^N \left[ (Na_i)^2 - 2Na_i + 1 \right] \geq 0$. However, I'm not sure how to prove the inequality after expanding.

How would you prove (or disprove) the above proposition?

• What happened when you expanded $\sum(Na_i-1)^2$? – Did Apr 28 '17 at 12:56
• @Did I got $\sum \left[ (Na_i)^2 - 2Na_i + 1 \right] \geq 0$ and I wasn't sure where to go from there. – Dando18 Apr 28 '17 at 12:59
• Why the downvote? Any advice is welcome. – Dando18 Apr 28 '17 at 13:00
• And this is how far you could go expanding? What about $\sum(Na_i)^2=N^2\sum a_i^2$? And $\sum2Na_i=2N\sum a_i=\ldots$? And $\sum 1=\ldots$? – Did Apr 28 '17 at 14:10

## 3 Answers

Cauchy-Schwarz applies here exactly like it does in your link: $$\left(\sum_{i=1}^N1^2\right)\left(\sum_{i=1}^Na_i^2\right)\geq\left(\sum_{i=1}^Na_i\right)^2=1^2=1\implies \sum_{i=1}^Na_i^2\geq\frac{1}{N}.$$

The same geometric approach I proposed in my previous answer applies here, too. The constraint $x_1+\ldots+x_N=1$ implies that the point $(x_1,\ldots,x_N)$ lies in a affine hyperplane $\pi$. Such hyperplane is orthogonal to the line $x_1=\ldots=x_N$, hence the point $P\in\pi$ minimizing the distance from the origin is the point with coordinates $x_1=\ldots=x_N=\frac{1}{N}$. As a consequence, for any $Q\in \pi$ the squared distance from the origin is $\geq N\cdot\left(\frac{1}{N}\right)^2 = \frac{1}{N}$, i.e. $$x_1^2+\ldots+x_N^2 \geq \frac{1}{N}$$ as wanted.

Because $\sum\limits_{i=1}^{N}a_i^2\geq\frac{1}{N}\left(\sum\limits_{i=1}^Na_i\right)^2$ it's just $\sum\limits_{1\leq i<j\leq N}(a_i-a_j)^2\geq0$.