Let $P(x)=ax^3+bx^2+cx+d$ be a polynomial with real coefficients such that \[\min\{d,b+d\}> \max\{|{c}|,|{a+c}|\}\] Prove that $P(x)$ do not have a real root in $[-1,1]$.
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Tags: algebra, polynomial, inequalities, algebra unsolved
30.04.2010 17:15
one way to solve this is case-working ... and the cases that we must consider is the sign of $a,b,c$ ... for example $a \ge 0 ,b \le 0 , c \le 0$ and so on on .... though this is a long solution but it's easy ...but there exists also a nice straightforward solution ...
05.05.2010 14:31
Yes, what I did was just chasing all 8 cases. No difficulty arose at all. Can you please post that straightforward solution?
09.05.2010 17:34
lasha wrote: Yes, what I did was just chasing all 8 cases. No difficulty arose at all. Can you please post that straightforward solution? it is a solution of my teacher, not mine: suppose that there exists a root x which: $-1\le{x}\le{1}$; so $0=\mid{ax^3+bx^2+cx+d}\mid\ge{\mid{bx^2+d}\mid-\mid{ax^3+cx}\mid}$ $=(b+d)x^2+d(1-x^2)-\mid {x((a+c)x^2+c(1-x^2))}\mid$ $\ge{(b+d)x^2+d(1-x^2)-\mid {(a+c)x^2+c(1-x^2)}\mid}$ $\ge{(b+d)x^2+d(1-x^2)- (\mid a+c\mid x^2+\mid c \mid (1-x^2))}$ $\ge{min(b+d,d)(x^2+1-x^2)-max(\mid{c}\mid,\mid{a+c}\mid)(x^2+1-x^2)}\gneq {0}$ contradiction.
10.05.2010 04:30
ArefS wrote: $\ge{min(b+d,d)(x^2+1-x^2)-max(\mid{c}\mid,\mid{a+c}\mid)(x^2+1-x^2)}\gneq {0}$ contradiction. I think the last inequality should be no "equal" sign.
10.05.2010 06:43
Justhalf wrote: ArefS wrote: $\ge{min(b+d,d)(x^2+1-x^2)-max(\mid{c}\mid,\mid{a+c}\mid)(x^2+1-x^2)}\gneq {0}$ contradiction. I think the last inequality should be no "equal" sign. If you look better, you'll see that the actual sign is "greater than and NOT equal to" (the bottom line is ticked over).
20.05.2010 16:46
I think it is better than chasing all 8 cases
29.03.2022 18:09
Let assume $-1 \le k \le 1$ such that $P(k) = (a+c)k^3 + (b+d)k^2 + ck(1-k^2) + d(1-k^2) = 0$. Note that $d > |{c}| > c \ge ck \implies (d+ck) > 0$ and $b+d > |{a+c}| > a+c \ge (a+c)k \implies (b+d)+(a+c)k) > 0$. $(a+c)k^3 + (b+d)k^2 \ge ((b+d)+(a+c)k)k^2 \ge 0$ and equality holds if $k = 0$. $ck(1-k^2) + d(1-k^2) = (d + ck)(1-k^2) \ge 0$ and equality holds if $k = +1$ or $-1$. So there exists no such $k$.
07.11.2023 09:33
Assume it does have a root $k$ satisfying this condition. Then, note that $|bk^2+d| = |k||ak^2+c|$. Then, if $b$ has the opposite sign as $d$, then by taking $k^2$ till 1, we decrease the value of $|bk^2+d|$, and if $b$ has the same sign as $d$, we can take $k^2$ to $0$. Thus, we get $|bk^2+d| \geq \min{d, b+d}$ (note that $d$ and $b+d$ are both nonnegative). Now similarly we can get $|ak^2+c| \leq \max{|c|, |a+c|}$. But now using $|k| \leq 1$, we get $\min{d, b+d} \leq \max{|c|, |a+c|}$, contradiction.
07.11.2023 12:30
Amir Hossein wrote: Let $P(x)=ax^3+bx^2+cx+d$ be a polynomial with real coefficients such that \[\min\{d,b+d\}> \max\{|{c}|,|{a+c}|\}\]Prove that $P(x)$ do not have a real root in $[-1,1]$. See also here