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- Feb 14, 2012

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For $x,\,y,\,z\ge 0$ and $x+y+z=1$, prove that $0\le xy+yz+zx-2xyz\le\dfrac{7}{27}$.

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- Thread starter
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- #1

- Feb 14, 2012

- 3,941

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For $x,\,y,\,z\ge 0$ and $x+y+z=1$, prove that $0\le xy+yz+zx-2xyz\le\dfrac{7}{27}$.

-----

Remember to read the POTW submission guidelines to find out how to submit your answers!

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- Feb 14, 2012

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Suggested solution from other:

It is easy to see that $xy+yz+zx-2xyz=xy(1-2z)+(x+y)z\ge 0$.

$\begin{align*}xy+yz+zx-2xyz&=xy(1-2z)+(x+y)z\\& \le \left(\dfrac{x+y}{2}\right)^2(1-2z)+(x+y)z\\&=2t\left(\dfrac{1}{4}+\dfrac{t}{2}\right)^2+\dfrac{1}{4}-t^2\\&=\dfrac{4t^3-4t^2+t-2}{8}\\&=f(t)\end{align*}$.

Let $f'(t)=0$ and we get $t=\dfrac{1}{6}$ and $t=\dfrac{1}{2}$.

$f''\left(\dfrac{1}{6}\right)=-7.5<0$ tells us the maximum of $f$ is $f\left(\dfrac{1}{6}\right)=\dfrac{7}{27}$.

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