How Does the Cauchy-Schwarz Inequality Prove a Vector Inequality?

[SeannyBoi71]

Homework Statement

Let u = [a b] and v = [1 1]. Use the Cauchy-Schwarz inequality to show that (a+b/2)² ≤ a²+b²/2. Those vectors are supposed to be in column form.

Homework Equations

|<u,v>| ≤||u|| ||v||,
and the fact that inner product here is defined by dot product (so <u,v> = u\cdotv)

The Attempt at a Solution

|<u,v>| ≤ ||u|| ||v||

|<[a b],[1 1]>| ≤ ||[a b]|| ||[1 1]||

|a+b| ≤ √(a²+b²)√2

and there is where I'm stuck. Any help please?

 

[Quinzio]

I can't understand... are a, b real numbers ? Any value ?
Take a=b=1 you have (\frac{3}{2})^2 &lt; \frac{3}{2}

 

[SeannyBoi71]

It doesn't specify that they are real numbers, but I can only assume they are supposed to be... and I don't think I can take specific numbers. I have to show the general case keeping a and b in there.

 

[YpouTalk]

Hey, after going through the question i found the answer (if you still need it/ anypone else needs to see it)

You have |x [dot] y| <= ||x|| [dot] ||y||

so: |<a,b>[dot]<1,1>| <= ||<a,b>|| [dot] ||<1,1>||

simplify to: |a+b| <= sqrt(a^2+b^2)*sqrt(2)

then square both sides to get: (a+b)^2 <= (a^2+b^2)*2
(you can see triangle inequality here)

Next, simplify other equation.

((a+b)/2)^2 <= (a^2 + b^2)/2

simplify to: (a+b)^2/4 <= (a^2 + b^2)/2

multiply both sides by 4: (a+b)^2 <= (a^2+b^2)*2

And you have both equations the same, therefore it holds.

Hope that helps