I have an interesting problem, in which i would be utmost happy to receive some help:
$p$, $q$, $r$ are real numbers with the following property: $|px^2+qx+r|\le 1$ for all $x$ in $[-1; 1]$. Prove that $|rx^2+qx+p|\le2$ for all $x$ in $[-1; 1]$.
Substituting for $x=0$ yields $-1\le r\le1$. $x=1$ and $x=-1$ results $-1\le p+q+r\le1$ and $-1\le p-q+r\le 1$ respectively. From those: $-2\le p\le2$ and $-1\le q\le 1$.
I couldn't go anything further. Would be happy to receive some ideas. :)
Let $P(x)=px^2+qx+r$ and $P_1(x)=rx^2+qx+p$.
W.L.O.G. we can assume $p\geq0$ because if not then we can consider $-P(x)=-(px^2+qx+r)$ satisfying all the given conditions and here the the leading coef. of $x^2$ is +ve.
W.L.O.G. we can also consider $q\geq0$ beacuse if not then we can talk abt $P(-x)$ satisfying all the given conditions.
Now putting $x=1,0,-1$ in turn we get,
$|p+q+r|\leq1,|r|\leq1,|p-q+r|\le1$
From this applying $\Delta$ inequalities we get,
$|r|\leq1,|p+q|\leq2,|p-q|\le2$
And further if $r\geq0$ then we have for $|x|\le1$,
$P_1(x)=rx^2+qx+p\leq r+q+p\le 1$ and $P_1(x)=rx^2+qx+p\geq 0+q(-1)+p\ge -2$
So it follws that $|P_1(x)|\leq 2$
Now if $r\leq 0$ then we have for $|x|\le1$,
$P_1(x)=rx^2+qx+p\leq 0+q+p\leq 2$ and $p_1(x)\ge r+q(-1)+p\ge p-q+r\ge-1$
So in this case also we have $|P_1(x)|\leq 2$
So we are done.