How to prove $\sum_{I=1}a_{I}^{2}$ is larger than $\frac{1}{n}$

Question

How to prove $\sum_{I=1}^{n}a_{I}^{2}$ given constraints that $\sum_{I=1}^{n}a_{I}$ = 1 is larger than or equal to $\sum_{I=1}^{n}(\frac{1}{n})^{2}$?

Answer 1

hint: Use the inequality: $(1\cdot a_1+1\cdot a_2+\cdots +1\cdot a_n)^2 \le (1^2+1^2+\cdots+1^2)(a_1^2+a_2^2+\cdots +a_n^2)$ known as Cauchy-Schwarz inequality.