Let $n\in\mathbb N$ and let $x_1,\dots,x_n>0$ be $n$ positive real numbers so that $x_1+\dots+x_n=1$. Use the Cauchy-Schwarz inequality and show that
a. $\sum_{i=1}^n x_i^2\geqslant \frac{1}{n}$.
b. $\sum_{i=1}^n \frac{1}{x_i}\geqslant n^2$.
Hint: Think about the dot product on $\mathbb R^n$. Then, choose specific vectors $\mathbf x,\mathbf y\in\mathbb R^n$ and apply the Cauchy-Schwarz inequality.
Solution:
a. Let $\mathbf x=[x_1,\dots,x_n]^\top$ and $\mathbf y=[1,\dots,1]^\top \in\mathbb R^n$. Then $$\langle \mathbf x, \mathbf y\rangle=x_1+\dots+x_n=1,$$ $$\langle \mathbf x, \mathbf x\rangle = x_1^2+\dots+x_n^2,$$ $$\langle \mathbf y, \mathbf y\rangle = 1+\dots+1=n.$$Applying the Cauchy-Schwarz inequality, we have $$\langle \mathbf x, \mathbf x\rangle\langle \mathbf y, \mathbf y\rangle\geqslant (\langle \mathbf x, \mathbf y\rangle)^2,$$which implies that $$n\sum_{i=1}^n x_i^2\geqslant 1.$$Therefore, we obtain $\sum_{i=1}^n x_i^2\geqslant \frac{1}{n}$.
b. Let $\mathbf x=[\sqrt{x_1},\dots,\sqrt{x_n}]^\top$ and $\mathbf y=[1/\sqrt{x_1},\dots,1/\sqrt{x_n}]^\top \in\mathbb R^n$. Then $$\langle \mathbf x, \mathbf y\rangle = 1+\dots+1=n,$$ $$\langle \mathbf x, \mathbf x\rangle = x_1+\dots+x_n=1,$$ $$\langle \mathbf y, \mathbf y\rangle =\frac{1}{x_1}+\cdots+\frac{1}{x_n}.$$Applying the Cauchy-Schwarz inequality, we have $$\langle \mathbf x, \mathbf x\rangle\langle \mathbf y, \mathbf y\rangle\geqslant (\langle \mathbf x, \mathbf y\rangle)^2,$$which implies that $$1\sum_{i=1}^n\dfrac{1}{x_i}\geqslant n^2.$$Therefore, we obtain $\sum_{i=1}^n \frac{1}{x_i}\geqslant n^2$.