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Problem Of The Week #394 Nov 27th, 2019

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
Here is this week's POTW:

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$x^3+ax^2+bx+c$ has three distinct real roots, but $(x^2+x+2001)^3+a(x^2+x+2001)^2+b(x^2+x+2001)+c$ has no real roots. Show that $2001^3+a(2001^2)+b(2001)+c>\dfrac{1}{64}$.

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

MHB POTW Director
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Feb 14, 2012
3,894
Congratulations to lfdahl for his correct solution, which you can find below:

Let $P(x) = (x^2+x+2001 - \alpha)(x^2+x+2001 - \beta) (x^2+x+2001 - \gamma)$, where $\alpha, \beta$ and

$\gamma$ are the three distinct real roots of $x^3+ax^2+bx+c$.

$P$ has no real roots, so $\alpha, \beta$ and $\gamma$ must all obey the inequality: $1^2-4(2001-r) < 0$

or $r < 2000\frac{3}{4}$. This is the minimum value of the polynomium $x^2+x+2001$ in $x = -\frac{1}{2}$.

Since $P(0) = (2001 - \alpha)(2001 - \beta) (2001 - \gamma) = 2001^3+a\cdot2001^2+b\cdot2001+c$, is a

function of the coefficients (i.e. the roots), and every root obeys $2001-r > 2001-2000\frac{3}{4} = \frac{1}{4}$

we get a sharp limit for $P(0)$:

$$P(0) > \left ( \frac{1}{4}\right )^3=\frac{1}{64}.$$
 
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