# Cauchy–Schwarz inequality

1. Feb 16, 2013

### VertexOperator

1. The problem statement, all variables and given/known data

Prove that:

2. Relevant equations

$$\sum_{k=1}^{n}x_{k}^{2}\geq \frac{1}{n}\left ( \sum_{k=1}^{n}x_{k} \right )^{2}$$

3. The attempt at a solution

I am not sure what to do to be honest. But it looks like the Cauchy–Schwarz inequality to me.

2. Feb 16, 2013

### micromass

Staff Emeritus
Re: Proof

Oh come on. I'm sure there are things you can try. For example, solve the case n=2. Or try induction. Or something.

3. Feb 16, 2013

### VertexOperator

Re: Proof

Sigma is intimidating :(
But I will try doing it.

4. Feb 16, 2013

### micromass

Staff Emeritus
Re: Proof

Yes, I agree that it is intimidating. That is why I usually do some special cases first like n=2 and n=3 to get a feeling for the proof!

5. Feb 16, 2013

### VertexOperator

Re: Proof

I got $$2x_{1}^{2}+2x_{2}^{2}\geq x_{1}^{2}+2x_{1}x_{2}+x_{2}^{2}$$

6. Feb 16, 2013

### micromass

Staff Emeritus
Re: Proof

Do you see why this is true?

7. Feb 16, 2013

### VertexOperator

Re: Proof

Is it because $$\left ( x_{1}^{2}+x_{2}^{2} \right )\geq 2x_{1}x_{2}$$
But why is this true?
PS: Is it because $$\left ( x_{1}-x_{2} \right )^{2}\geq 0 \therefore \left ( x_{1}^{2}+x_{2}^{2} \right )\geq 2x_{1}x_{2}$$

8. Feb 16, 2013

### VertexOperator

Re: Proof

Do I just do normal induction from here?

9. Feb 16, 2013

### micromass

Staff Emeritus
Re: Proof

After thinking a little bit about it, you won't need induction at all.

Basically, you need to prove that

$$\sum_{k=1}^n x_k \leq \sqrt{n}\sqrt{\sum_{k=1}^n x_k^2}$$

You were completely right that this resembled Cauchy-Schwarz. So, try to apply Cauchy-Schwarz somewhere.

Of course, for that inequality, you need two sequences. The sequence $(x_1,...,x_n)$ is one of them, what is the other?

10. Feb 16, 2013

### VertexOperator

Re: Proof

Is it $$x_{1}^{2},...,x_{n}^{2}$$ ?

And thank you a lot for the help. I didn't think about squarerooting both sides :( I really need to learn to recognized this stuff.

11. Feb 16, 2013

### micromass

Staff Emeritus
Re: Proof

No, it's not that.

Think of a very easy (constant) sequence.

12. Feb 17, 2013

### VertexOperator

Re: Proof

Although this question can be done using the CS Inequality, it can be done this way too:
$$a^2+b^2 \geq 2ab$$

[We need to sub into a and b, all the combinations of 2 numbers from the set $$\ \ x_1, \ x_2, \ x_2 \ \dots, \ x_n$$

$$x_1^2+x_1^2 \geq 2x_1 x_2$$

$$x_1^2+x_3^2 \geq 2x_1x_3$$

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.
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$$x_1^2+x_n^2 \geq 2x_1x_n$$

$$x_2^2+x_3^2 \geq 2x_2 x_3$$

And so on, then we add these inequalities side by side.

$$(n-1)(x_1^2+x_2^2 \dots + x_n^2) \geq 2(x_1x_2+x_1x_3+ \dots)$$

$$n(x_1^2+x_2^2+\dots+x_n^2) \geq x_1^2+x_2^2+\dots + x_n^2 + 2(x_1x_2+\dots$$

The RHS is the expanded form of:

$$n\sum_{k=1}^{n} x_k^2 \geq \left(\sum_{k=1}^{n}x_k \right)^2$$

$$\sum_{k=1}^{n} x_k^2 \geq \frac{1}{n}\left(\sum_{k=1}^{n}x_k \right)^2$$

13. Feb 17, 2013

### VertexOperator

Re: Proof

Could you please show me how you would do it using the CS inequality?

14. Feb 17, 2013

### micromass

Staff Emeritus
Re: Proof

You need to show that

$$\sum_{k=1}^n x_k \leq \sqrt{n}\sqrt{\sum_{k=1}^n x_k^2}$$

The CS inequality gives something like

$$\sum_{k=1}^n x_ky_k \leq ...$$

If I use the same $x_k$ in both inequalities above, then what does the corresponding $y_k$ need to be?

15. Feb 17, 2013

### VertexOperator

Re: Proof

Ok, thank you. I will try to do it now.