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Problem Of The Week 46

with 3 comments


Let be distinct positive reals. Determine the minimum value of \frac{1}{(x-y)^2}+\frac{16}{xy}+6(x+y) and determine when that minimum occurs.

Source: ALtheman’s problem column

Written by Implex

October 7, 2008 at 9:22 pm

3 Responses

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  1. Now first looking at the terms, the first thing I instinctively did was, I took y=x+1..so now the equation reduces to 1+16/(x*(x+1))+6(2x+1)..The function will be increasing in the interval (0,1) and after 1. So the minimum is obtained at x=1 which is 27

    Celebrating Life

    October 8, 2008 at 6:43 am

  2. Oops interchanged x and y..I considered 1/(y-x)^2

    Celebrating Life

    October 8, 2008 at 6:48 am

  3. minimum value occurs..when x=2 y=1 or y=2 x=1…
    the expression equals 27…for any other positive value the expression is greater than this….
    Let x=ky
    1/(y^2(k-1)^2)+ 16/(K*y^2)+ 6y(k+1)
    we can try substituing values for K and Y….

    I dont think this is best approach…just trial and error

    Ankaj

    October 8, 2008 at 2:37 pm


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