The Mathematical Proof Thread

[Kernel Sohcahtoa]

Hello everyone.  I was working on a proof yesterday and thought that it was cool enough to share here. Hence, the proposition and my direct proof of it is below.

With that said, in creating this thread, it was my intent for people to be able to post whatever beauty they encountered in mathematics for as long as possible. Thus, I apologize if my post violates community guidelines or causes any inconvenience to the staff and/or AF members. 

Proposition: For every four real numbers a,b,c, and d, sqrt(a^2 + b^2)*sqrt(c^2 + d^2) ≥ ac + bd.

Note the proposition can be re-written in “if-then” form, which yields the following equivalent statement: “if a,b,c, and d are real numbers, then sqrt(a^2 + b^2)*sqrt(c^2 + d^2) ≥ ac + bd.
 
Proof.

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Let a,b,c, and d be real numbers.  Then ad-bc is also a real number.  Now, the square of any real number is non-negative, so (ad-bc)^2 ≥ 0  (1). Expanding the left-side of inequality (1) gives (ad)^2 - 2abcd + (bc)^2 ≥ 0 (2).  Adding the real numbers (bd)^2, (ac)^2, and 2abcd to both sides of inequality (2) gives (ac)^2 + (ad)^2 + (bc)^2 + (bd)^2  ≥ (ac)^2 + 2abcd +(bd)^2, and rewriting the left-hand side of the foregoing inequality via exponential properties yields, a^2c^2 + a^2d^2 + b^2c^2  + b^2d^2  ≥ (ac)^2 + 2abcd +(bd)^2 (3).  Factoring both sides of inequality (3) gives (a^2 + b^2)*(c^2 + d^2) ≥ (ac + bd)^2 (4).  Taking the square root of both sides of inequality (4) yields sqrt((a^2 + b^2)*(c^2 + d^2)) ≥ sqrt((ac + bd)^2), or equivalently (via properties of radicals) , sqrt(a^2 + b^2)*sqrt(c^2 + d^2) ≥ ac + bd.  Hence, the proof is complete.

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Notes

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[1] The analysis of proof for this proposition was pretty cool.  Essentially, I worked backwards from the conclusion that sqrt(a^2 + b^2)*sqrt(c^2 + d^2) ≥ ac + bd, and via algebra, I ended up reaching the statement that (ad-bc)^2 ≥0 , which was my last statement in the backward process.  As a result, I worked forward from the hypothesis that a,b,c, and d are real numbers. I then generated the following forward statements: (1) ad-bc is a real number; (2) the square of any real number is non-negative, so (ad-bc)^2 ≥0.  Thus, the last statement in the forward process is precisely the last statement in the backward process, which indicated to me that I had a proof.

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[2]This proposition can also be proven via the contradiction method.  To proceed with contradiction, we would take the negation of the conclusion that sqrt(a^2 + b^2)*sqrt(c^2 + d^2) ≥ ac + bd, which is sqrt(a^2 + b^2)*sqrt(c^2 + d^2) < ac + bd. We then work forward from this new statement and the hypothesis in order to generate a contradiction. Also, if you work backward from my steps in the direct proof (taking into account the different inequality sign), then via algebra, you should be able to reach the following contradiction:

(ad-bc)^2 < 0.

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