Question

Let P(x) be a polynomial with non-negative coefficients. Prove that if the inequality P(1/x)P(x)≥1 holds for x=1, then this inequality holds for all positive x.

Sorry for reposting. I think the first one got lost in the afternoon torrent of questions. Happy valentines!

Answer 1

If this is true for x = 1, then what means is
\[ P(1)^2 \geq 1 \]

Answer 2

I think I'm missing something obvious, but how did you come to that conclusion?

Answer 3

(Incoming facepalm moment, I'm sure...)

Answer 4

Now consider the function f(x) = P(1/x)P(x)

One way to approach this problem would be to show that for x > 1 , f'(x) > 0 and for 0 < x < 1, f'(x) < 0. In which case, 1 is a local and in fact global minimum of f(x) and therefore

\[ f(x) \geq 1 \ \ \] for all real numbers.

Answer 5

Aha! Alright, I see.

Answer 6

See if that works for you. I haven't worked this through yet, but I imagine it's a good strategy to start with.

Answer 7

Oh, I'm totally off. For some reason I solved for P(x)+P(1/x) instead of P(x)P(1/x). I will try again :-)

Answer 8

So we are given that P(x) has all positive coefficients,
\[ P(x) = \sum_{i=0}^n a_ix^i \] where \( a_i \geq 0 \).

Now it's clear that for \( P'(x) > 0 \) for \(x > 1 \) and that \( P'(1/x) < 0 \) for \( x > 1 \) for similar reasons.

So far so good?

Answer 9

Sorry, JamesJ, I'm not seeing it. :( Also, I get the feeling that MM's "wrong" proof was still the right way to go about it.

Answer 10

Lol, near simultaneous posts.

Answer 11

How do we know that a is greater than or equal to 0?

Answer 12

"Let P(x) be a polynomial with non-negative coefficients"

Answer 13

Woops. My bad.

Answer 14

Sorry, this is what happens when I try to multitask. :| Alright, so far so good.

Answer 15

Now
\[ f'(x) = P(1/x)P'(x) - \frac{1}{x^2}P'(1/x)P(x) \]
Show now that for \( x > 1 \), \( f'(x) > 0 \)

Answer 16

And in the meantime, if anyone has a more elegant proof, the floor is yours

Answer 17

@James: I think your proof doesn't include the interval (0,1), does it?

Answer 18

No, it doesn't. But show this part and then you can get the other by (anti)symmetry.

Answer 19

Ah I see.

Answer 20

I get it so far. But what do you mean by "antisymmetry"?

Answer 21

I think he meant something like this: for every number \(a>1\), there exist a number \(b<1\) such that \(P(a)P(\frac{1}{a})=P(b)P(\frac{1}{b})\), namely \(b=\frac{1}{a}\).

Answer 22

Huh. While I mull over the implications of that, I'd like to point out the "elegant" proof for this is as attached.

Answer 23

I didn't come up with it, naturally. XD

Answer 24

Right.

Answer 25

Solving it without Cauchy would be cool too, though.

Answer 26

I don't really know this inequality. So for me I would have to prove this inequality first.

Answer 27

I'm sure you do it, just that it's disguised here. The usual form is this: consider any two vectors v and w in \( \mathbb{R}^n \). Then
\[ u \cdot v \leq ||u|| \ ||v|| \]

Answer 28

Man, I'm confused. What course am I supposed to learn this in?

Answer 29

Oh, wait, nevermind. I'm not confused. XD I type faster than I think.

Answer 30

lol.

Answer 31

You're meta-confused: confused about your state of confusion.

Answer 32

Haha.

Answer 33

Oh, by the way, the non-Cauchy solution as given. Attached. Naturally, again, I was not involved in doing this. :P Although this one makes a lot more sense to me.

Answer 34

Now that I think about it, though, that middle looks an awful lot like Cauchy...

Answer 35

This last proof is what I've been trying to do. I like it more than the others.

Answer 36

For what it's worth, to finish off my first strategy:
If P(1)^2 = 1, then P(1) = 1 (as all the coefficients are positive) and P(x) > 0 for all x > 1; similarly P(1/x) < 1 for all x > 1.

Therefore
f'(x) = P'(x)P(1/x) - (1/x^2)P(x)P'(1/x) > P'(x).1 - 1.1.P'(1/x) > 0

Now, for x > 1, 0 < 1/x < 1, and
\[ (f(1/x))' = -(1/x^2)f'(1/x) = -(1/x^2)f'(x) \] because f(x) = f(1/x) Now that last expression is less than zero because f'(x) > 0 for x > 1.

Hence f has the properties we want and hence the inequality is proved.

===

The C-S proof is still better. Much better. And no, there second proof doesn't use it, but another calculus result instead: that for any positive integer n
\[ x^n + \frac{1}{x^n} \geq 2 \]

Answer 37

*their second proof

Answer 38

I found your proof the easiest to understand. :P Thanks for the help. I was hoping to derive another method of proving it, other than the two I provided.