(Exercise 1.1.8 in Real Analysis 2 by Tao) Show that
$$\sum_{i=1}^n |x_i - y_i] \le \sqrt{n}\sqrt{\sum_{i=1}^n (x_i -y_i)^2}.$$
The book suggests to use either Cauchy or triangle inequality, but I don't know how to use this. I appreciate if you give some help.
On
Cauchy-Schwarz inequality, states that:
$$\left(\sum_{i=1}^n a_ib_i \right)^2\leq \left(\sum_{i=1}^na_i^2\right)\left(\sum_{i=1}^nb_i^2\right)$$
and in particular for $b_i=1$:
$$\left(\sum_{i=1}^n a_i \right)^2\leq n\left(\sum_{i=1}^na_i^2\right)$$
Given the form of the question, what values should we choose for $a_i$ ?
Square both sides. The equivalent formulation of the problem is now to show
$$\left( \sum_{i=1}^n |x_i-y_i| \right)^2 \le n \sum_{i=1}^n (x_i-y_i)^2$$
In the typical formulation of Cauchy-Schwarz for sums,
$$\left(\sum_{k = 1}^{n} a_k b_k\right)^2 \leq \left(\sum_{k=1}^{n}a_k^2\right) \left(\sum_{k=1}^{n}b_k^2\right)$$
take $a_k$ to be your $|x_i - y_i|$ and take $b_k = 1$. Some nice simplification should follow.