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Thread: Proving a tricky inequality

  1. #1
    Feb 2010

    Question Proving a tricky inequality

    Hi, this might be very easy but I am unable to prove it.

    (From now on, the summations will be defined over the set {R,P,M})

    Let $\displaystyle \phi =\phi _{R}\alpha _{R}+\phi _{M}\alpha _{M}+\phi _{P}\alpha _{P}$ ; where

    $\displaystyle \forall i\in \left \{ R,P,M \right \} \phi _{i}\in (0,\infty )$
    and $\displaystyle \alpha _{i}\in (0,1 )$ and $\displaystyle \sum \alpha _{i}=1$
    and $\displaystyle y_{i}\in (0,\infty )$

    and also, define $\displaystyle y=\sum \alpha _{i}y_{i}$

    Now the claim is:

    1) If $\displaystyle \phi _{R}> \phi > \phi _{P}$; then $\displaystyle \frac{\sum \alpha _{i}\phi _{i}y_{i}}{\phi }> y$
    2) If $\displaystyle \phi _{R}< \phi < \phi _{P}$; then $\displaystyle \frac{\sum \alpha _{i}\phi _{i}y_{i}}{\phi }< y$

    Alpha's being in between 0 and 1 and summing to 1 seems to be crucial for this fact, since I tried around 50 examples in excel and it holds if thats the case.

    Any help is appreciated.

    p.s. forgot to add, we also have
    $\displaystyle y_{R}> y_{M}>y_{P}$

    edit: I got it, it's not tricky at all actually. If anyone's interested, just post a reply here and I'll write the solution.
    Last edited by ichoosetonotchoosetochoos; Jan 15th 2012 at 09:34 AM.
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