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Math Help - A Proof for the Inequality of Arithmetic and Geometric Means

  1. #1
    Senior Member bkarpuz's Avatar
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    A Proof for the Inequality of Arithmetic and Geometric Means

    Dear friends,

    while working on some computations, I figured out that my arguments supply a proof for
    the arithmetic and geometric mean inequality by using calculus. I wanted to share here.

    Assertion. For all x,y\geq0 show that \dfrac{x+y}{2}\geq\sqrt{xy} with equality if and only if x=y.
    Proof. The case xy=0 is trivial.
    Let x,y>0 and define
    f(x,y):=\dfrac{x+y}{2}-\sqrt{xy}.
    We shall prove that for all x,y>0, f(x,y)\geq0 with equality if and only if x=y.
    In this case, we may find \lambda>0 such that y=\lambda x (in particular, \lambda:=y/x), which yields
    f(x,\lambda x)=g(\lambda)x, where g(\lambda):=\dfrac{\lambda+1}{2}-\sqrt{\lambda} for \lambda>0.
    As x>0, we have to show that for all \lambda>0, g(\lambda)\geq0 with equality if and only if \lambda=1.
    We can compute that
    g^{\prime}(\lambda)=\dfrac{1}{2}\bigg(1-\dfrac{1}{\sqrt{\lambda}}\bigg) and g^{\prime\prime}(\lambda)=\dfrac{1}{4\sqrt{\lambda  ^{3}}}>0 for all \lambda>0.
    The function g is therefore convex and this shows that g(1)=0 is the global minimum value
    since g^{\prime}(1)=0. Finally, for all \lambda>0, g(\lambda)\geq g(1)=0 with equality if and only if \lambda=1. \rule{0.2cm}{0.2cm}
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  2. #2
    MHF Contributor chisigma's Avatar
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    An alternative is the use of polar coordinates...

    x= \rho\ \cos \theta

    y= \rho\ \sin \theta (1)

    ... so that is...

    \displaystyle \frac{x+y}{2} = \frac{\rho}{2} (\sin \theta + \cos \theta)

    \displaystyle \sqrt{x\ y} = \rho\ \sqrt{\sin \theta \ \cos \theta} (2)

    Now the proof of the 'inequality' is equivalent to prove that is...

    \displaystyle \frac{1}{4} \ge \frac{\sin \theta \ \cos \theta}{2} (3)

    ... and (3) is true because the function f(\theta) = \sin \theta \ \cos \theta has its maximum for \theta= \frac{\pi}{4} and is f(\frac{\pi}{4}) = \frac{1}{2}...

    Kind regards

    \chi \sigma
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    Here is an easy proof.
    It should be clear that (a-b)^2\ge 0.
    So that a^2+b^2\ge 2ab.
    If x\ge 0~\&~y\ge 0 then let a=\sqrt{x}~\&~ a=\sqrt{y} .
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by chisigma View Post
    \displaystyle \frac{1}{4} \ge \frac{\sin \theta \ \cos \theta}{2} (3)
    \displaystyle\frac{\sin (\theta )\cos \theta }{2}\le \frac{1}{4}\implies \sin (2\theta )\le 1.
    Last edited by Krizalid; November 9th 2010 at 01:51 PM.
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  5. #5
    Senior Member bkarpuz's Avatar
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    @bkarpuz: I should take \lambda^{2} instead of \lambda.
    @chisigma: This makes use of \lambda=\tan(\theta).
    @Plato: I know that that is the most famous and easiest proof. ^^
    @Krizalid: Nice trick!
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