# Proving Inequalities Involving Summations and Sq. Roots

How can I prove that $$\sum_{i=1}^{n} |a_i| \leq \sqrt{n} \sqrt{\sum_{i=1}^{n} a_{i}^{2}}$$ considering that $$a_{1}, a_{2}, a_{3}, ... , a_{n}$$ are real numbers?

This exercise was presented in a section that also covered Cauchy Schwarz: $$(\sum_{i=1}^{n}a_{i}b_{i})^2 \leq (\sum_{i=1}^{n}a_{i}^2)(\sum_{i=1}^{n}b_{i}^2)$$ but I am unsure if anything related to Cauchy Schwarz is involved in this proof.

On the RHS, I proceeded with $$\sqrt{n} \sqrt{\sum_{i=1}^{n} a_{i}^{2}}$$ = $$\sqrt{n\sum_{i=1}^{n} a_{i}^{2} }$$ but then I'm stuck and unsure how to proceed. Any hints/help in this direction is greatly appreciated.

Thank you!

• Write $|a_i| = 1\cdot |a_i|$ and apply C-S. – Song Jan 11 at 21:44

Let $$b_i =1$$ for $$1\le i\le n$$
Then apply Cauchy Schwartz inequality to $$|a_i|$$ and your $$b_i=1$$
It's $$\sqrt{(1^2+1^2+...+1^2)(a_1^2+a_2^2+...+a_n^2)}\geq$$ $$\geq\sqrt{1^2\cdot a_1^2}+\sqrt{1^2\cdot a_2^2}+...\sqrt{1^2\cdot a_n^2}=|a_1|+|a_2|+...+|a_n|.$$