# Maybe triangle inequalities theorem

1. Aug 2, 2012

### azizlwl

1. The problem statement, all variables and given/known data
Prove: If a2+b2=1, and c2+d2=1, then ac+bd ≤1

2. Relevant equations

Maybe triangle inequalities theorem.

3. The attempt at a solution

2. Aug 2, 2012

### gabbagabbahey

Re: Inequalities

For this question to make sense, I think a,b,c and d must be Real numbers. Is this given?

Hint: What is (a+c)2+(b+d)2?

3. Aug 2, 2012

### azizlwl

Re: Inequalities

Thanks
At the beginning of the chapter, "An inequalities is a statement that one(Real)number is greater than or less than another".
Is there inequalites in complex number?

(a+c)2+(b+d)2≥0
equal zero if both factors=0, greater than zero if one or both factor not equal to zero.
a2+c2+2ac +b2+d2+2bd≥0
2+2ac+2bd≥0
ac+bd≤-1 ???

So it should be
(a-c)2+(b-d)2≥0
Equal to zero if a=c and b=d, greater than zero if one or both of components of the factors not equal.
a2+c2-2ac +b2+d2-2bd≥0
2-2ac-2bd≥0
ac+bd≤1

Last edited: Aug 2, 2012
4. Aug 2, 2012

### gabbagabbahey

Re: Inequalities

One can extend the definition of an inequality to deal with complex numbers (by comparing their real and imaginary parts), but this isn't nessecary in this case since if $a$ and $c$ are both purely imaginary, their product will still be real (but the inequality that you are supposed to prove may not be satisfied).

Anyways, if you are studying real numbers then it is probably safe to assume all 4 numbers (a,b,c & d) are real.

This is true, but not all that useful here. instead of looking at the lower limit of (a+c)2+(b+d)2, use the triangle inequality (twice) to look at its upper limit.

Keep in mind, that if x2+y2=1 (for real-valued x & y), then |x|^2=x^2=1-y^2≤1

This method also works and is even simpler than mine!

5. Aug 3, 2012

### HallsofIvy

Staff Emeritus
Re: Inequalities

But (not relevant to this question but just so that other readers do not get the wrong idea) that extension of inequality does not make the complex numbers an ordered field. In particular, if x< y and 0< z, it is NOT necessarily true that xz< yz.