- #1
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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