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 ...

Yes, what I did was just chasing all 8 cases. No difficulty arose at all. Can you please post that straightforward solution?

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.

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.

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).

I think it is better than chasing all 8 cases

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 $k$.

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.

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