The Mathematical Proof Thread

[Kernel Sohcahtoa]

Thanks for your last post, Polymath.  I must admit that I kept getting stuck during each of my attempts and decided to let it go for the time being.  However, this evening, I noticed that your post also appeared as an exercise in one of my math books, so I decided to give it another try. 

That said, for anyone interested, I will post the result to be proved and then post the proof in hide tags.  Also, in my proof, I make use of another fact/result, so I will post that fact and my proof of it (in hide tags) above my main proof. That said, should anyone read the proofs below, please feel free to point out any errors/typos that I have made; I'm very appreciative of any feedback that you can offer.  Thanks

Result: for every two positive real numbers a and b, it follows that a/b +b/a ≥ 2. 

Proof

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Suppose a and b are positive real numbers.  Then (a–b)^2 ≥ 0 which is equivalent to a^2 –2ab +b^2 ≥ 0 which is equivalent to a^2 + b^2 ≥ 2ab.  Dividing both sides of the foregoing inequality by ab (recall that a and b are both positive reals, so ab is positive) gives (a^2 + b^2)/ab ≥ 2 which is equivalent to a^2/ab + b^2/ab ≥ 2 which is equivalent to a/b + b/a ≥ 2.  The proof is complete.   

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Main Result to Prove: for every n ≥ 1 positive real numbers a_1,a_2,…,a_n, it follows  that (a_1 +…+ a_n)*(1/a_1 +…+ 1/a_n) ≥ n^2

Proof

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The proof will be by induction.  For n=1, we have a_1*(1/a_1) = 1  ≥ 1^2=1, which is true.  Now, assume that (a_1 +…+ a_k)*(1/a_1 +…+ 1/a_k) ≥ k^2 where k ≥ 1.  We must establish that (a_1 +…+ a_k+1)*(1/a_1 +…+ 1/a_k+1) ≥ (k+1)^2.  To that end, observe that

(a_1 +…+ a_k+1)*(1/a_1 +…+ 1/a_k+1) = [(a_1 +…+ a_k) + a_k+1]*[(1/a_1 +…+ 1/a_k) + 1/a_k+1] = (a_1 +…+ a_k)*(1/a_1 +…+ 1/a_k) + [(a_1 +…+ a_k)*1/a_k+1] + [(1/a_1 +…+ 1/a_k)*a_k+1] + (a_k+1)*(1/a_k+1)= (a_1 +…+ a_k)*(1/a_1 +…+ 1/a_k) + [(a_1 +…+ a_k)*1/a_k+1] + [(1/a_1 +…+ 1/a_k)*a_k+1] + 1 ≥ k^2 + [(a_1 +…+ a_k)*1/a_k+1] + [(1/a_1 +…+ 1/a_k)*a_k+1] + 1 .



Thus, so far, we have obtained the inequality (a_1 +…+ a_k+1)*(1/a_1 +…+ 1/a_k+1) ≥ k^2 + [(a_1 +…+ a_k)*1/a_k+1] + [(1/a_1 +…+ 1/a_k)*a_k+1] + 1 (1) (note that the result in inequality (1) follows from the induction hypothesis).



Now, let’s focus on the expressions [(a_1 +…+ a_k)*1/a_k+1] and [(1/a_1 +…+ 1/a_k)*a_k+1] in inequality (1).  We can rewrite each expression as follows: rewrite [(a_1 +…+ a_k)*1/a_k+1] as (a_1/a_k+1 +…+ a_k/a_k+1) (2)  and [(1/a_1 +…+ 1/a_k)*a_k+1 as (a_k+1/a_1 +…+ a_k+1/a_k) (3).  Now adding the first term in  expression (2) with the first term in expression (3) followed by adding the second term in expression (2) with the second term in expression (3) and continuing this addition up to the kth terms in expression (2) and expression (3) gives

[(a_1/a_k+1) + (a_k+1/a_1)] + [(a_2/a_k+1) + (a_k+1/a_2)] + …+ [(a_k/a_k+1) + (a_k+1/a_k)] (4). Now, for every two positive real numbers a and b, it follows that a/b + b/a ≥ 2.  Now, using the first sum in expression (4) as an example, since a_1 and a_k+1 are two positive real numbers, then it follows that (a_1/a_k+1) + (a_k+1/a_1) ≥ 2; thus, for any two positive real numbers a_i and a_k+1, where  1≤ i  ≤ k, it follows that (a_i/a_k+1) + (a_k+1/a_i) ≥ 2.  Consequently, adding each (a_i/a_k+1) with each corresponding (a_k+1/a_i), which is how expression (4) is set up, yields k sums, and so , it follows that [(a_1/a_k+1) + (a_k+1/a_1)] +…+ [(a_k/a_k+1) + (a_k+1/a_k)] ≥ 2k. So, we have established that [(a_1 +…+ a_k)*1/a_k+1] + [(1/a_1 +…+ 1/a_k)*a_k+1] = [(a_1/a_k+1) + (a_k+1/a_1)] +…+ [(a_k/a_k+1) + (a_k+1/a_k)] ≥ 2k (5).


Now, via result (5), it follows from inequality (1) that (a_1 +…+ a_k+1)*(1/a_1 +…+ 1/a_k+1) ≥ k^2 + [(a_1 +…+ a_k)*1/a_k+1] + [(1/a_1 +…+ 1/a_k)*a_k+1] + 1 ≥ k^2 + 2k +1= (k+1)^2.


Hence, by the principle of mathematical induction, for every n ≥ 1 positive real numbers a_1, a_2,…, a_n, it follows that (a_1 +…+ a_n)*(1/a_1 +…+ 1/a_n) ≥ n^2.  The proof is complete. 

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