# Proving an inequality (not induction)

1. Nov 3, 2008

### Bleys

1. The problem statement, all variables and given/known data
For any real numbers a,b,c,d, prove that
$$\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)\geq\left(ac+bd\right)^{2}$$

2. The attempt at a solution
I use the triangle inequality to show that
$$\left|ac+bd\right|^{2}\leq\left(\left|ac\right|+\left|bd\right|\right)^{2}$$

$$\left(ac\right)^{2}+\left(bd\right)^{2}+2abcd\leq\left(ac\right)^{2}+\left(bd\right)^{2}+2\left|ac\right|\left|bd\right|$$

But I'm not sure how to compare it with the final result. Can you assume that
$$2\left|ab\right|\left|cd\right|\leq\left(bc\right)^{2}+\left(ad\right)^{2}$$
I know the answer is probably no, and you have to take cases of a<b<c<d, a<c<b<d, etc. But that's 24 cases to consider! And while some are analogous to others, it doesn't cut it by more than half.

Am I even remotely on the right track? Any help or hint would be appreciated.

2. Nov 3, 2008

### Staff: Mentor

You might try a proof by contradiction.
Assume that (a^2 + b^2)(c^2 + d^2) < (ac + bd)^2

Expand both sides and move the terms on the right side over to the left. If you can end up with the square of something on the left side being less than zero, there's your contradiction.

3. Nov 3, 2008

### tiny-tim

Hi Bleys!
Nope.

Hint: forget triangle inequalities …

expand both sides (as Mark44 says), and then look at the difference …

does it remind you of something?

4. Nov 3, 2008

### Bleys

Oh right; so (ad-bc)^2 < 0, so the starting inequality must be true;
I did think about proof by contradiction but it was to prove
$$2\left|ab\right|\left|cd\right|\leq\left(bc\right) ^{2}+\left(ad\right)^{2}$$
-_- feel kinda silly now..

Thank you for help, both you!