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Math Help - How can this inequality be proved?

  1. #1
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    How can this inequality be proved?

    If x, y > 0 and 0 < p < 1, then (x+y)^p < x^p+y^p.
    It was on Wikipedia:
    http://en.wikipedia.org/wiki/Inequality_(mathematics)#Power_inequalities
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  2. #2
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    Re: How can this inequality be proved?

    Quote Originally Posted by Karima View Post
    If x, y > 0 and 0 < p < 1, then (x+y)^p < x^p+y^p.
    It was on Wikipedia:
    http://en.wikipedia.org/wiki/Inequality_(mathematics)#Power_inequalities
    There are a few fun ways to prove this.

    First, consider the case
    \, x=y.
    Note then that
    (x+y)^p = (x+x)^p = (2x)^p = 2^p(x^p)
    and
     x^p + y^p = x^p + x^p = 2(x^p).
    Since
    0<p<1,
    it follows that
    2^p<2,
    and therefore,
     (x+y)^p < x^p+y^p when x=y.

    Next, assume, Without Loss of Generality, that \, x>y.

    (x+y)^2 = \left(x \left(1 + \tfrac{y}{x} \right)\right)^p = x^p \left(1+\tfrac{y}{x}\right)^p

    Since \tfrac{y}{x}<1, we have an expansion of the form
    (1+r)^p, 0<r<1.

    I'm a bit tired, so perhaps you can finish this or I will later on.
    Thanks from Karima
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  3. #3
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    Re: How can this inequality be proved?

    Let me rephrase your inequality so what follows is more clear: (a+b)^p<a^p+b^p where a,b>0 and 0<p<1.
    If a=b, 2^pa^p<2a^p since 2^p<2 -- the last inequality since the function x^p is increasing.
    So suppose a\ne b. I can suppose a<b. (The problem is symmetric in a and b.) Consider the function f(x)=(x+1)^p-x^p on (0,\infty). Now f^{\prime}(x)<0 --
    (x+1)^{p-1}-x^{p-1}<0 since x^{p-1} is decreasing. Since \lim_{x\rightarrow 0^+}f(x)=1, the function f can be extended to a continuous function g on [0,\infty) where g(0)=1 and g(x)=f(x) otherwise. Now g is decreasing on [0,\infty) -- the proof that a function on [0,\infty), which has a negative derivative in the open interval (0,\infty), uses the mean value theorem on [0,\infty); remember that differentiability is required only on the open interval. So for any x>0, 1=g(0)>g(x)=(x+1)^p-x^p for any x>0 or (x+1)^p<x^p+1
    Finally, (a/b+1)^p<(a/b)^p+1 and easily then (a+b)^p<a^p+b^p.
    Thanks from Karima
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