Proof this inequality using Chebyshev's sum inequality

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playboy

Homework Statement



Let a,b,c,d,e be positive real numbers. Show that

[tex]\displaystyle{\frac{a}{b+c}} + \displaystyle{\frac{b}{c+d}} + \displaystyle{\frac{c}{d+e}} + \displaystyle{\frac{d}{e+a}} +\displaystyle{\frac{e}{a+b}} \geq \displaystyle{\frac{5}{2}} [/tex]



Homework Equations



Chebyshev's sum inequality:
http://en.wikipedia.org/wiki/Chebyshev's_sum_inequality



The Attempt at a Solution



Assume: a > b > c > d > e

Then: a+b > a+e > b+c > c+d > d+e

Or: [tex] \displaystyle{\frac{1}{d+e}} \geq \displaystyle{\frac{1}{c+d}} \geq \displaystyle{\frac{1}{b+c}} \geq \displaystyle{\frac{1}{a+e}} \geq \displaystyle{\frac{1}{a+b}} [/tex]


Hence, we have:

[itex] a \geq b \geq c \geq d \geq e \geq[/itex]

[tex] \displaystyle{\frac{1}{d+e}} \geq \displaystyle{\frac{1}{c+d}} \geq \displaystyle{\frac{1}{b+c}} \geq \displaystyle{\frac{1}{a+e}} \geq \displaystyle{\frac{1}{a+b}} [/tex]


But...this [tex]\sum a_{k}*b_{k}[/tex] is NOT matching up like it what the question is asking...

Am I arranging these numbers wrong?


What I am trying to say is:


[tex] \displaystyle{\frac{a}{d+e}} + \displaystyle{\frac{b}{c+d}} + \displaystyle{\frac{c}{b+c}} + \displaystyle{\frac{d}{a+e}} + \displaystyle{\frac{e}{a+b}} [/tex]

IS NOT EQUAL TO

[tex]\displaystyle{\frac{a}{b+c}} + \displaystyle{\frac{b}{c+d}} + \displaystyle{\frac{c}{d+e}} + \displaystyle{\frac{d}{e+a}} +\displaystyle{\frac{e}{a+b}} \geq \displaystyle{\frac{5}{2}} [/tex]
 
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Answers and Replies

  • #2
AKG
Science Advisor
Homework Helper
2,565
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First, you can't assume a > b > c > d > e, since the sum on the left is not invariant under arbitrary permutations (it is invariant under cyclic permutations (abcde)k though). Also, how are you getting a+e > b+c? That's not a valid inference. And yes, the thing you end up with doesn't match up anyways.
 

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