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Math Help - inequalities

  1. #1
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    inequalities

    If  x+y = 3z prove that x^2 + y^2 >= 3z^2

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  2. #2
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    Note that 3z^2=3(3z/3)^2, so one has to show that x^2+y^2\ge3((x+y)/3)^2. After multiplying both sides by 3 and moving everything to the left, that expression can be represented as a sum of two squares.
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  3. #3
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    Quote Originally Posted by emakarov View Post
    Note that 3z^2=3(3z/3)^2, so one has to show that x^2+y^2\ge3((x+y)/3)^2. After multiplying both sides by 3 and moving everything to the left, that expression can be represented as a sum of two squares.
    Maybe I have a blind spot, but I don't see that last step of getting to the sum of two squares. Is there some identity that I've forgotten ? .

    I can get to the result using the AM >= GM inequality, (so x^{2}+y^{2}\geq 2xy).
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  4. #4
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    \begin{aligned}<br />
x^2+y^2&\ge3((x+y)/3)^2\Leftrightarrow{}\\<br />
3x^2+3y^2-x^2-2xy-y^2&\ge0\Leftrightarrow{}\\<br />
x^2-xy+y^2&\ge0\Leftrightarrow{}\\<br />
(x-y/2)^2+3y^2/4&\ge0<br />
\end{aligned}<br />
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  5. #5
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    Nice one. Thanks, I didn't see that .
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