# Method checking to show the given inequality to be true

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?

• 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. May 12 at 19:09
• Understood Sir thanks May 13 at 4:05

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.

• Nice approach thanks Sammy Black 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.

• 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 May 13 at 4:06