# Thread: triangle ineqaulity

1. ## triangle ineqaulity

How I prove the triangle ineqaulity by the inequalty of Cauchy-Swartz?

2. ## Re: triangle ineqaulity Originally Posted by policer How I prove the triangle ineqaulity by the inequalty of Cauchy-Swartz?
What are you calling the inequalty of Cauchy-Swartz?
There are several forms of inequalities that go by some form of that name.
Please reply.

4. ## Re: triangle ineqaulity Originally Posted by policer Well, in that link you are given proofs for several forms to the inequality.
What more do you require?

5. ## Re: triangle ineqaulity

I will give you an answer tomorrow I hope.

6. ## Re: triangle ineqaulity Originally Posted by policer I will give you an answer tomorrow I hope.
You understand that we do not have a copy of the notes/textbook from which you are working?

Here is a classic proof if one is working with absolute value as a norm.

7. ## Re: triangle ineqaulity

I can't solve the problem with Wikipedia.
To solve it I need this form:
sqrt(a^2+b^2)/(c^2+d^2) => ac + bd
the zigzag line abc is a broken line between ab.
ab is a straight line.
Tell me so, how to formula it?

8. ## Re: triangle ineqaulity Originally Posted by policer To solve it I need this form:
sqrt(a^2+b^2)/(c^2+d^2) => ac + bd
the zigzag line abc is a broken line between ab.
ab is a straight line
.
I have absolutely no idea what any of that means.

Suppose that each of $a~\&~b$ is a real number.
\begin{align*}|a|\cdot |b|&=|a\cdot b| \\&\ge a \cdot b \end{align*} so that $2|a|\cdot |b|\ge 2a\cdot b$.

\begin{align*}|a+b|^2&=(a+b)^2 \\&=a^2+2a\cdot b+b^2\\&\le |a|^2+2|a\cdot b|+b^2\\ &\le a^2+2|a|\cdot |b|+|b|^2\\&\le (|a|+|b|)^2 \end{align*}
Using the square root we get the result.