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Thread: Inequality

  1. #16
    Lord of certain Rings
    Isomorphism's Avatar
    Dec 2007
    IISc, Bangalore
    That is a rather neat solution

    Quote Originally Posted by JaneBennet View Post
    By the way, some people think this problem is easy.

    Art of Problem Solving Forum

    Unfortunately I don’t understand the steps in their solution.
    I assume you did not understand this:
    \frac {a^2}{a^3 - abc + a} + \frac {b^2}{b^3 - abc + b} + \frac {c^2}{c^3 - abc + c}\ge \frac {(a + b + c)^2}{a^3 + b^3 + c^3 - 3abc + a + b + c}$

    Try proving $\displaystyle \frac{x^2}{a} +\frac{y^2}{b} +\frac{z^2}{c} \geq \frac{(x+y+z)^2}{a+b+c}$
    With a little bit of algebra, you will see this is equivalent to,

    $\displaystyle \sum_{cyc}{x^2(\frac{b+c}{a})} \geq 2 \sum_{cyc}{xy}$
    Which after a little bit of bunching shows clear AM-GM.

    EDIT: Actually it directly follows from CS(in a slightly different form)
    $\displaystyle \frac{x_{1}^{2}}{a_{1}}+\frac{x_{2}^{2}}{a_{2}}+.. .+\frac{x_{n}^{2}}{a_{n}}\geq\frac{(x_{1}+x_{2}+.. .+x_{n})^{2}}{a_{1}+a_{2}+...+a_{n}}$
    where $\displaystyle x_{i}\in\mathbb{R}$ and $\displaystyle a_{i}\geq0$
    Last edited by Isomorphism; Dec 29th 2007 at 08:53 PM. Reason: Added CS solution
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