Maybe triangle inequalities theorem

[azizlwl]

Homework Statement

Prove: If a²+b²=1, and c²+d²=1, then ac+bd ≤1

Homework Equations

Maybe triangle inequalities theorem.

The Attempt at a Solution

 

[gabbagabbahey]

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?

 

[azizlwl]

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?

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

 

[gabbagabbahey]

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?

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.

a²+c²+2ac +b²+d²+2bd≥0
2+2ac+2bd≥0
ac+bd≤-1 ?

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

So it should be
(a-c)²+(b-d)²≥0
a²+c²-2ac +b²+d²-2bd≥0
2-2ac-2bd≥0
ac+bd≤1

This method also works and is even simpler than mine!

 

[HallsofIvy]

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).

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.

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!