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Thread: Inequality holds?

  1. #1
    Jun 2010

    Inequality holds?

    I'm sorry to repost this thread, but I put it in the pre-univ algebra section and I had no suggestions.

    There's a proof I need to complete which requires that the following inequality holds:

    $\displaystyle \frac{1}{4}((b-a_1)^2+(b-a_2)^2+(1-a_3)^2+(1-a_4)^2) \geq$ $\displaystyle \frac{1}{6}((b-a_1)^2+(b-a_2)(2\cdot b-a_1-a_2)+(b-a_3)(2\cdot b-a_4-a_3)+(b-a_4)^2)$


    $\displaystyle 0\leq a_1\leq a_2\leq a_3\leq a_4\leq b\leq 1$

    I have made several plots and in all of them the inequality holds. Nonetheless, I need to finish this problem theoretically. I have already made the proof for the case of $\displaystyle a_1=a_2\neq a_3=a_4$, and also for the case where $\displaystyle a_1=a_2=a_3=a_4$, but these are straightforward, and the real deal comes when I want to solve for the general case.

    I know $\displaystyle 0\leq (2\cdot b-a_4-a_3)\leq (2\cdot b-a_1-a_2)\leq 2$. I also know that $\displaystyle (b-a_3)(2\cdot b-a_4-a_3)\leq (b-a_2)(2\cdot b-a_1-a_2)$. Finally, I also know that the case where both sides of the inequality are maximums occurs when $\displaystyle a_1=a_2=a_3=a_4=0$ and $\displaystyle b=1$.

    I hope somebody here can help me with this problem. Best regards.
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  2. #2
    MHF Contributor Swlabr's Avatar
    May 2009
    With a bit of work (take everything to one side, collect terms, multiply by 2 and then spot the 6 elements which can be collected into two squares) you can show what you want if you can show that,

    $\displaystyle 1+a_3b+a_4b \geq b^2+a_3+a_4$.

    This inequality holds (can you see why?).

    I apologise, but I cannot seem to find a `neat' solution to this...
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  3. #3
    Jun 2010
    Sorry Swlabr, but I'm totally lost.

    I put everything on one side and collect? I cannot find the expression you want me to see.

    I have tried a different approach, but I don't know if it is correct. I first found the limit (moving everything to the left side) when $\displaystyle b\rightarrow1$. The solution is:

    $\displaystyle (a_2-a_1)^2+(a_4-a_3)^2\geq 0$

    This holds, but I don't know if this is sufficient. If I find the limit when $\displaystyle b\rightarrow 0$, I also know that $\displaystyle a_i\rightarrow 0$, which gives me:

    $\displaystyle 6\geq 0$

    which trivially holds. If,finally, $\displaystyle b\rightarrow a_4$:

    $\displaystyle (a_4+2a_3-3)^2+(a_3-3)^2-3+(a_2-a_1)^2-\frac{1}{4}(4a_3-6)^2\geq 0$

    From this I know:

    $\displaystyle (a_3-3)^2-3\geq 0$

    but I don't know if the rest holds. Nonetheless, can I use this kind of argument?
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  4. #4
    Jun 2010


    As you can see from the original definition, left side of the inequality is the Euclidean distance and right hand is a distance function I have defined. From definition of a distance function, it will be monotonically increasing, so I think that by making the proof in the extremes (when b is either 0 or 1) also proves for the general case. I mean, if a distance function is smaller in the extremes of the domain than the Euclidean function, there is no way this relationship will change in the rest of the domain, as both are monotonically increasing. What do you think?
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