Given quadratic polynomial $f(x)$ satisfies $\lvert ax^2 + bx + c \rvert \leq \lvert x \rvert$ for all $x \in [-1,1]$. Show that $\lvert a \rvert + \lvert b \rvert \leq 1$.

My approach was graphical , when we make a quadratic and try satisfying the given inequality, we get few points to notice from the graph itself. Suppose $a \geq 0$ then we will have a upward opening parabola. For it to satisfy the inequality that is graph of |quadratic| to lie below the $\lvert x \rvert$ graph on $[-1,1]$.

(1) It must have equal roots in $[-1,1]$ (that is, it should be touching the $x$-axis only), consider it intersects at two points on $x$-axis then the graph of |quadratic| will have that negative curve to be reflected above, making it always intersect the $|x|$ twice on $[-1,1]$. But that we don't want to.

(2) We need the equal roots to occur at $x=0$. If not then because of symmetric nature about the minima point of the parabola, we would be having in one part that is either $[0, 1]$ or $[-1, 0]$ to be satisfied by the $\lvert x \rvert$ inequality, but in other half its not or partially satisfied in both regions.

From this we can conclude $f(x) = +x^2$ as the possible solution to it. Same we can argue for $a \leq 0$ we would get $f(x) = -x^2$ as possible solution.

Another case would be of a linear polynomial that is degree one, then in that case by similar analogy we can say only $f(x) = \pm x$ satisfy it. From these all we can observe $\lvert a \rvert + \lvert b \rvert = 1$ is being satisfied so and less than case would be the $f(x) = cx$ or $cx^2$ where $-1 < c < 1$? Hence proved? If its all correct please confirm and any another method after checking whether this all is correct can be also appreciated.

Curious thought: can this graphical approach be applicable for cubic, fourth degree ones? If there were some past research being done in this method, can it be shown here?

  • 2
    $\begingroup$ Very interesting if somewhat bizarre posting. The problem is directly conquered by analysis. That is, after concluding that $c = 0$, you know that for any two real numbers $r$ and $s$, if they are both positive or both negative, then $|r + s| = |r| + |s|$. So, examining $x = 1, x = -1$ conquers the problem. Therefore, the underlying question is whether you can construct a similar problem, involving a higher degree polynomial, that is more readily conquered by graphical examination rather than analysis. $\endgroup$ May 12 at 19:09
  • $\begingroup$ Understood Sir thanks $\endgroup$ May 13 at 4:05

2 Answers 2


You can establish this inequality, by only evaluating the polynomial at $x \in \{-1, 0, 1\}$: \begin{align} x &= -1 &&\Longrightarrow &\lvert a - b \rvert &\leq 1 \tag{$-1$}\\ x &= 0 &&\Longrightarrow &\lvert c \rvert &\leq 0 \tag{$0$}\\ x &= 1 &&\Longrightarrow &\lvert a + b \rvert &\leq 1 \tag{$1$} \end{align}

Immediately from Equation $(0)$, we see that $c=0$.

Since the polynomial is quadratic, $a \neq 0$. If $b=0$, then $\lvert b \rvert = 0$, and by Equation $(1)$, $$ \lvert a \rvert + \lvert b \rvert = \lvert a \rvert = \lvert a + b \rvert \leq 1, $$ as desired.

Now, assume that $b \neq 0$. There are two cases: $ab > 0$ (same signs) or $ab < 0$ (opposite signs).

If the coefficients have same signs, then $\lvert a \rvert + \lvert b \rvert = \lvert a + b \rvert$, so by Equation $(1)$,
$$ \lvert a \rvert + \lvert b \rvert = \lvert a + b \rvert \leq 1. $$ On the other hand, if the coefficients have opposite signs, then $\lvert a \rvert + \lvert b \rvert = \lvert a - b \rvert$, so by Equation $(-1)$,
$$ \lvert a \rvert + \lvert b \rvert = \lvert a - b \rvert \leq 1. $$

Can you prove the equations used in the final two paragraphs?

Lemma. For any two nonzero real numbers $a$ and $b$:

  • If $a$ and $b$ have same signs, then $\lvert a \rvert + \lvert b \rvert = \lvert a + b \rvert$.
  • If $a$ and $b$ have opposite signs, then $\lvert a \rvert + \lvert b \rvert = \lvert a - b \rvert$.

Hint: there are $4$ cases to consider.

  • $\begingroup$ Nice approach thanks Sammy Black $\endgroup$ May 13 at 4:07

I'll describe a very different approach.

Firstly, $x=0 \implies |c| \leq 0 \implies c=0$.

Now, $x=1 \implies |a+b| \leq 1$ and $x=-1 \implies |-a+b| = |a-b| \leq 1$.

Now, $|a|=\pm a, |b|=\pm b$. So, it's easy to see (just check $4$ possible cases) that $|a|+|b|$ is either $|a+b|$ whenever $ab \geq 0$ and $|a-b|$ whenever $ab \leq 0$.

So, we're done!

Just noticed that @user2661923 has made a very similar comment. If you want to post it as an answer, I'm happy to delete mine.

  • $\begingroup$ No its fine as both are in all solutions to the problem , i do not mind if both have the same ideas :) thanks for this $\endgroup$ May 13 at 4:06

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