- #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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