Prove Minkowski Inequality using Cauchy-Schwartz Inequality

[Rederick]

I expanded (x+y),(x+y) and got x^2+y^2 > 2xy then replaced 2xy with 2|x,y| but now I'm stuck.

I need to get it to ||x+y|| <= ||x|| + ||y||. Am I close?

 

[Rederick]

Here's what I did so far...

Let x=(x1,x2..xn) and y=(y1,y2..yn) in R. Assume x,y not = 0. Then (x+y)dot(x+y) = sum(x^2+2xy+y^2) >= 0. Then I rewrote it as sum(x^2 +y^2) >= sum(2xy). Using Cauchy Schwartz Inequality, sum(2xy) = 2|x dot y| <=2( ||x|| ||y||). So now I have this:

sum(x^2) +sum(y^2) <= 2( ||x|| ||y||).

I'm not even sure I'm doing the right thing. Can anyone help?

 

[Fredrik]

Start with $\|x+y\|^2$. This is less than what?

I'm assuming that what you're trying to prove is that

$$|\langle x,y\rangle|\leq \|x\|\|y\|\Rightarrow \|x+y\|\leq\|x\|+\|y\|$$

I'm not familiar with the term "Minkowski inequality". I would call the inequality on the right the "triangle inequality". (Edit: Aha, it's the triangle inequality for a specific Hilbert space).

By the way, if you click the quote button next to this post, you can see how I did the LaTeX. Keep in mind that there's a bug that causes the wrong images to appear in previews most of the time, so you will need to refresh and resend after each preview.

 

[Rederick]

Thank you Fredrik. I got it.

 

[blue_raver22]

Yes, you are on the right track. To prove the Minkowski inequality using the Cauchy-Schwartz inequality, we first expand (x+y)^2 as (x+y)(x+y)=x^2+2xy+y^2. Then, using the Cauchy-Schwartz inequality, we can rewrite 2xy as 2|x||y|. Substituting this into the expanded expression, we get (x+y)^2 = x^2 + 2|x||y| + y^2. Now, we can take the square root of both sides to get ||x+y|| = √(x^2 + 2|x||y| + y^2).

Next, we can use the triangle inequality, which states that for any two vectors x and y, ||x+y|| ≤ ||x|| + ||y||. Applying this to our expression, we get √(x^2 + 2|x||y| + y^2) ≤ √(x^2) + √(2|x||y|) + √(y^2). Since √(x^2) = ||x|| and √(y^2) = ||y||, we can rewrite this as ||x+y|| ≤ ||x|| + √(2|x||y|) + ||y||.

Now, we can use the fact that √(2|x||y|) ≤ √(2||x||^2 + 2||y||^2) = √(2(||x||^2 + ||y||^2)). This is a direct result of the Cauchy-Schwartz inequality. Substituting this into our previous expression, we get ||x+y|| ≤ ||x|| + √(2(||x||^2 + ||y||^2)) + ||y||. Finally, we can use the fact that √(a+b) ≤ √a + √b for any positive real numbers a and b to simplify this expression further. This gives us ||x+y|| ≤ ||x|| + ||y|| + √(2(||x||^2 + ||y||^2)).

Therefore, we have shown that ||x+y|| ≤ ||x|| + ||y|| + √(2(||x||^2 + ||y||^2)), which is