# Prove the Triangle Inequality

blanik
I have proven the triangle inequality starting with ||a+b||^2 and using the Schwartz Inequality. However, the next part of the problem says:

"Show that the Triangle Inequality is an equality if and only if |a>=alpha|b> where alpha is a real positive scalar." It must be proved in both directions.

Any help on where to begin would be greatly appreciated.

## Answers and Replies

fourier jr
blanik said:
I have proven the triangle inequality starting with ||a+b||^2 and using the Schwartz Inequality. However, the next part of the problem says:

"Show that the Triangle Inequality is an equality if and only if |a>=alpha|b> where alpha is a real positive scalar." It must be proved in both directions.

Any help on where to begin would be greatly appreciated.

begin by proving one direction. complete the proof by proving the other direction. i'm not sure what |a>=alpha|b> means, is there another way to explain what that says?

blanik
|a>=alpha|b> means the vector A equals alpha times the vector B where alpha is a real positive scalar. Does that help?

I understand that I am "supposed" to start with one way and go the other, but what does that mean? Do I substitute a=alpha b for a and solve for ||alpha b + b|| = ||alpha b|| + ||b||? I have been playing around with the definition of ||a|| = SQRT (a a*), etc...

fourier jr
in your proof of the triangle inequality make all your inequalities equalities & see what you get. at the step where you use the cauchy-schwartz inequality you see that (a,b) = |a||b|. if one vector is a multiple of the other then figure out that (a,b) = |a||b| is true. for the other way suppose that's true. then by the cauchy-schwartz inequality one vector is a multiple of the other. ( ( , ) means inner product & | | means length)

Start by explicitly writing out what $$||~ |a \rangle + |b \rangle ~ ||^2$$ is. You might start seeing where the Scwartz inequality comes into play.