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December 17, 2017 at 4:39 pm (This post was last modified: December 17, 2017 at 4:41 pm by Kernel Sohcahtoa.)
Earlier this month, I did a proof that was pretty cool (IMO). As a result, I thought I'd share it here. As always, I'll put my post in hide tags. In addition, I'd like to clarify that my writing is not meant to be arrogant: my mathematical writing style is how I've learned to communicate mathematics (I definitely have lots of room for improvement). With that said, here is the following exercise from my mathematical proofs book by Gary Chartrand:
Prove that if a,b, and c are positive real numbers, then (a+b+c)(1/a + 1/b + 1/c) ≥ 9
Hint/Analysis of Proof
Let’s work backward from the conclusion and see where we end up: I will denote each backward statement with a B and a respective number. First, let’s notice that the expression (1/a + 1/b +1/c) can be re-written as (bc+ac+ab)/abc. Thus, the conclusion can be re-written as
Factoring the expressions in each bracket in B6 gives
B7: a(b – c)^2 + b(a – c)^2 + c(a – b)^2 ≥ 0.
Now, since a,b, and c are positive real numbers (via the hypothesis), we know that the inequality in B7 is non-negative.
Now, turning to the forward process, we can reach our last result in the backward process (B7) by making use of some properties of real numbers and the hypothesis that a,b, and c are positive real numbers (Note, I will label each statement in the forward process with an A and a respective number). To that end, since a,b, and c are real numbers (positive real numbers in this case), then
A1: b – c, a – c, and a – b are real numbers.
So, via the properties of real numbers, it follows that
A2: (b – c)^2, (a – c)^2, and (a – b)^2 are non-negative real numbers.
Since a,b, and c are positive real numbers, then
A3: a(b – c)^2, b(a – c)^2, and c(a – b)^2 are non-negative real numbers
Thus, it follows that
A4: a(b – c)^2 + b(a – c)^2 + c(a – b)^2 ≥ 0.
Hence, since we’ve reached our B7 statement, we have a proof, and writing the proof will consist of retracing our backward steps in a forward direction. Please note that, since we’ve included a lot of the fine details in our analysis, we can write a more condensed version of the proof.
Proof (condensed version)
Suppose a,b, and c are positive real numbers. Then b – c, a – c, and a – b are also real numbers, and by the properties of real numbers, (b – c)^2, (a – c)^2, and (a – b)^2 are non-negative real numbers. Consequently, it follows that a(b – c)^2 + b(a – c)^2 + c(a – b)^2 ≥ 0 (1). Now, expanding the left side of inequality (1) and using algebra gives [a(b^2) + a(c^2)] + [(a^2)b + b(c^2)] + [(a^2)c + (b^2)c] – 6abc ≥ 0 (2). Adding 9abc to both sides of inequality (2) and rearranging terms on the left side of inequality (2) gives (a^2)c + (a^2)b + (b^2)c + a(b^2) + b(c^2) + a(c^2) + 3abc = [abc + (a^2)c + (a^2)b] + [(b^2)c + abc + a(b^2)] + [b(c^2) + a(c^2) + abc] ≥ 9abc (3). Factoring the left side of inequality (3) yields
(a+b+c)(bc+ac+ab) ≥ 9abc (4). Finally, dividing the right side of inequality (4) by abc and dividing the expression (bc+ac+ab) on the left side of inequality (4) by abc gives (a+b+c)(1/a + 1/b + 1/c) ≥ 9, which is the desired result. Hence, the proof is complete.
P.S. I've included an analysis of proof in order to illustrate the thought processes that are involved in constructing a proof (please note that I had more scratch work for each of the statements listed in the analysis; thus, these steps were the result of me taking the time to work things out on paper). For me at least, I gain an understanding of a proof via the analysis of proof and not by the condensed version. Well, thanks for your time and attention, and I hope that more people will post cool math stuff in this thread. Also, there was no solution for this exercise in the book, so the work that I have posted here is entirely my own.