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Problem Of The Week #441 November 2nd, 2020

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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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Let $a,\,b,\,c$ be three real numbers such that $1\ge a \ge b \ge c \ge 0$. Prove that if $k$ is a (real or complex) root of the cubic equation $x^3+ax^2+bx+c=0$, then $|k|\le 1$.

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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
Staff member
Feb 14, 2012
3,894
Congratulations to lfdahl for his correct solution (Cool) , which you can find below:
Suppose $k$ (complex or real) is a root. And suppose, that $|k| > 1$.

By the triangle inequality we have:

$0 = \left | k^3 +ak^2+ bk+c\right | \leq \left | k^3 \right |+\left | ak^2 \right |+\left | bk \right |+|c| = \left | k \right |^3+a\left | k \right |^2+b\left | k \right |+\left | c \right |, \;\;\;\;\;(1)$

The right hand equality holds because $a,b,c \geq 0$. In addition, since $1 \geq a \geq b \geq c $, we know, that $\left | k \right |^3 > a\left | k \right |^2 > b\left | k \right | > \left | c \right | \geq 0$

Thus the RHS in $(1)$ is greater than or equal to zero. Hence the only possible value of $|k|$ is zero, which is a contradiction.

We can conclude, that $|k| \leq 1$.
 
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