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Math Help - Prove x^1/2 is concave

  1. #1
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    Prove x^1/2 is concave

    I have to prove this inequality to show that the function f(x)=x^{1/2} is concave.

    I am just stucked with the algebra

     (\mu x+ ( 1- \mu ) y )^{1/2} \geq \mu x^{1/2}+(1-\mu ) y^{1/2}

    For x>0,y>0, 0<mu<1
    Last edited by altave86; January 15th 2010 at 04:40 PM.
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  2. #2
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    Quote Originally Posted by altave86 View Post
    I have to prove this inequality to show that the function f(x)=x^{1/2} is concave.

    I am just stucked with the algebra

     (\mu x+ ( 1- \mu ) y )^{1/2} \geq \mu x^{1/2}+(1-\mu ) y^{1/2}

    For x>0,y>0, 0<mu<1

    First of all, note that x \geq 0.

    y = x^{\frac{1}{2}} = \sqrt{x}

    \frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}}

    \frac{d^2y}{dx^2} = -\frac{1}{4}x^{-\frac{3}{2}}

     = -\frac{1}{4\sqrt{x^3}}.


    The function does not have derivative at x = 0 (why), and for all x > 0, the second derivative is < 0. This means that the gradient is always decreasing. Hence, a concave function.
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  3. #3
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    Thank you!

    However, I was asked to prove concavity by using the inequality approach I specified in the OP.

    It is the same one to prove that a set is convex, but with the inequality reversed.
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