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Should be an easy proof

  1. Oct 1, 2008 #1


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    1. The problem statement, all variables and given/known data
    Show that for all positive a,b
    a+b <= a^2/b + b^2/a

    3. The attempt at a solution
    This simplifies to showing that
    (a+b)(ab) <= a^3 + b^3
    a^2b + b^2a <= a^3 + b^3

    But I'm not really sure where to go from here... any hint would be appreciated.
  2. jcsd
  3. Oct 2, 2008 #2
    It's Interesting that a+b = a^2/b + b^2/a when a=b.

    So I might presume that a+b < a^2/b + b^2/a when a not equal b.

    Try b = a + epsilon, and eliminate b.

    Apply it to a^2b + b^2a <= a^3 + b^3.
  4. Oct 2, 2008 #3


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    How does this look?

    As noted, if [tex] a = b [/tex] we have equality. Since the original expression is symmetric in the two variables, we can assume W.L.O.G. that [tex] a < b [/tex].

    a+b & \le \frac{a^2}{b} + \frac{b^2}{a} \\
    \intertext{if and only if}
    a^2b + ab^2 & \le a^3 + b^3 \\
    \intertext{if and only if}
    ab^2 - b^3 & \le a^3 - a^2 b \\
    b^2(a-b) & \le a^2 (a-b) \\
    \intertext{if and only if}
    b^2 & \le a^2

    and, since the square function is strictly increasing on the positives, the final statement is true if and only if [tex] a < b [/tex].

    Since equality holds when the two are equal, and inequality holds when they are not, we are done.
  5. Oct 3, 2008 #4

    Gib Z

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    Perhaps it would have been easier to start with nicksauce's idea:

    Try factoring the RHS.
  6. Oct 3, 2008 #5


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    Easier? Nay, infinitely easier (well, not infinitely, but much easier).
    That's the problem with proofing one's own work - you get wrapped up and can't see the trees for the forest. Good point Gib Z.
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