# Show using Cauchy-Schwarz inequality [duplicate]

Show for any real numbers $a_1,a_2,...,a_n$
$$(a_1+a_2+···+a_n)^2 \leq n(a_1^2+a_2^2+...+a_n^2)$$

I know the definition of Cauchy-Schwarz is $$(\sum_{i=1}^n a_ib_i)^2 \leq \sum_{i=1}^n a_i^2 \sum_{i=1}^n b_i^2$$ I can write the same problem with following form, but I would not know how to continue to prove it $$(\sum_{i=1}^n a_i)^2 \leq n(\sum_{i=1}^n a_i^2)$$

Note: an other definition of Cauchy Schwarz $$\langle\ v,u\rangle^2 \leq \langle\ u,u\rangle \langle\ v,v\rangle$$

Could you give me a steps what to use?

## marked as duplicate by Arnaud D., Community♦Sep 20 '18 at 21:29

Hint:

Choose $b_i$ wisely. It is a constant.

If it is not clear which constant to pick, just let it be an arbitrary non-zero constant.

Your statement doesn't look correct. Let n be 1 and $a_1$ be 1/2. Then the statement would say $1/2 \leq (1/2)^2$, which isn't true.

• did you left out the square on the LHS? – Siong Thye Goh May 17 '18 at 2:40
• I think that's new; an update since I originally posted. I'm on my phone though, so i could be wrong. I may have missed it. @Thye Goh – NicNic8 May 17 '18 at 4:19

I get an other solution using the second definition of Cauchy-Schwarz $$\langle u,v \rangle \leq \langle u,u \rangle \langle v,v \rangle$$

take $u = (1,1,...,1)$ as an vector of order n and be $v = (a_1,a_2,...,a_n)$

Then $\langle u,v \rangle = (a_1+a_2+...+a_n)$

$\langle u,u \rangle = \sqrt{n}$

$\langle v,v \rangle = \sqrt{a_1^2+a_2^2+...+a_n^2}$

If remove the square root we have: $$(a_1+a_2+...+a_n)^2 \leq n(a_1^2+a_2^2+...+a_n^2)$$