Is the following statement true? $\frac{1}{n} \leq \sum_{i=1}^{n}\alpha_i^2$ where $\sum_{i=1}^n \alpha_i = 1$

Question

Is the following statement true?:

$\frac{1}{n} \leq \sum_{i=1}^{n}\alpha_i^2$

where $\sum_{i=1}^n \alpha_i = 1$

My guess is yes, but i can't prove it mathematically. Any ideas?

Answer 1

Inequality Cauchy-Schwarz:

$$ (a_1^2+\cdots+a_n^2)(b_1^2+\cdots+b_n^2)\ge a_1b_1+\cdots+a_nb_n $$ Hence $$ n(a_1^2+\cdots+a_n^2)=(a_1^2+\cdots+a_n^2)(1^2+\cdots+1^2)\ge a_1\cdot 1+\cdots+a_n\cdot1=1 $$ and hence $$ a_1^2+\cdots+a_n^2\ge\frac{1}{n} $$