Proving Inequality: a1 + a2 + · · · + an−1 + an

In summary, The conversation discusses a mathematical inequality involving positive real numbers and how to prove it. One person suggests using the AM-GM inequality for the case n=2, while another person suggests using the Cauchy-Schwarz inequality for the general case. The conversation ends with a comment about the flu affecting one's thinking ability.
  • #1
hypermonkey2
102
0
hey guys, i came across this inequality in analysis and am not sure how to prove it. Any ideas? It's not homework, I am just curious..

Let a1, a2, . . . , an be strictly positive real numbers. Show that

a1 + a2 + · · · + an−1 + an <= ((a1)^2)/a2 + ((a2)^2)/a3) +...+ ((an)^2)/a1



cheers
 
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  • #2
Here's the case n=2. It follows from the AM-GM inequality.

[tex]
0 \leq \left(\frac{a_1}{\sqrt{a_2}} - \frac{a_2}{\sqrt{a_1}}\right)^2 = \frac{a_1^2}{a_2} - 2\sqrt{a_1 a_2} + \frac{a_2^2}{a_1} \leq \frac{a_1^2}{a_2} - (a_1 + a_2) + \frac{a_2^2}{a_1}.
[/tex]
 
  • #3
Thats pretty sweet. Thats hard to generalize, no?
 
  • #4
Yeah. I thought it was an easy induction argument after that, but that's obviously not the case. The flu is making me stupid.

Anyway, we can use the Cauchy-Schwarz inequality instead.

[tex]\left(\sum a_i\right)^2 = \left(\sum \frac{a_i}{\sqrt{a_{i+1}}} \sqrt{a_{i+1}}\right)^2 \leq \sum \frac{a_i^2}{a_{i+1}} \sum a_{i+1},[/tex]

where the index of summation is taken mod n (so a_{n+1} = a_1).
 

FAQ: Proving Inequality: a1 + a2 + · · · + an−1 + an

What is "Proving Inequality: a1 + a2 + · · · + an−1 + an"?

"Proving Inequality: a1 + a2 + · · · + an−1 + an" is a mathematical concept that involves proving the inequality between the sum of a series of numbers (a1, a2, ..., an−1, an) and another number or series of numbers.

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