# Employing Cauchy-Schwartz inequality question

Let $d_i\ge 0$ and $$\sum_{i=1}^n d_i=c$$ and $$\sum_{i=1}^nd_i^2\le C$$ Show that $$\frac{c^2}{n}\le C$$ The text says

Using Cauchy-Schwartz inequality and the first equation, show the last equaliton.

I do not know how to apply C-S to this. The inequality roughly states $$(u,v)^2\le(u,u)(v,v)$$ which is a trivial equality when $u=v$....

We have

$$(\sum_{i=1}^n d_i)^2\ge\sum_{i=1}^n d_i^2$$ but I dont know what to do with it...

By Cauchy-Schwarz Inequality you have:

$$(1 + 1 + \cdots + 1)(d_1^2 + d_2^2 + \cdots + d_n^2) \ge (d_1 + d_2 + \cdots + d_n)^2$$

Can you finish it now?

Hint: A version of CS inequality:

$$\left(\sum_{i=1}^nc_id_i\right)^2\leq\sum_{i=1}^nc_i^2\sum_{i=1}^nd_i^2$$

Hint: $$\left(\sum_{i=1}^nd_i\cdot1\right)^2\le\left(\sum_{i=1}^nd_i^2\right)\left(\sum_{i=1}^n1^2\right)$$

Using the root-mean square $\ge$ AM inequality (which can be derived from C-S):

$$\sqrt{\frac{C}{n}} \,\ge\, \sqrt{\frac{\sum_{i=1}^nd_i^2}{n}} \,\ge\, \frac{\sum_{i=1}^n d_i}{n} \,=\, \frac{c}{n}$$