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Let $P(x)$ be a polynomial with positive coefficients. Prove that if

$P \left (\dfrac{1}{x} \right ) \geq \dfrac{1}{P(x)}$

holds for $x=1$ , then it holds for all $x > 0$ .

#Sharky

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Let $n = \deg P$ and $\displaystyle P(x) = \sum_{i=0}^n a_ix^i$ .

First, note that if the inequality holds for $x = 1$ , then we have

$P(1) \geq \frac{1}{P(1)} \longrightarrow P(1)^2 = \left( \sum_{i=0}^n a_i \right)^2 \geq 1 \qquad (*)$

We recall the Cauchy-Schwarz inequality, which states that $\displaystyle \left( \sum_{i=0}^n \alpha_i^2 \right)\left( \sum_{i=0}^n \beta_i^2 \right) \geq \left( \sum_{i=0}^n \alpha_i\beta_i \right)^2$ . If we let $\alpha_i = \sqrt{a_ix^i}$ and $\beta_i = \sqrt{a_i/x^i}$ , then we have

$\left( \sum_{i=0}^n a_ix^i \right)\left( \sum_{i=0}^n \frac{a_i}{x^i} \right) \geq \left( \sum_{i=0}^n a_i \right)^2$

Now, we observe that $\displaystyle \sum_{i=0}^n a_ix^i = P(x)$ and $\displaystyle \sum_{i=0}^n \frac{a_i}{x^i} = P\left(\frac{1}{x}\right)$ . Using $(*)$ , we have

$P(x)P\left(\frac{1}{x}\right) \geq \left( \sum_{i=0}^n a_i \right)^2 \geq 1$

Hence, we divide both ends by $P(x)$ to obtain our desired equality:

$\boxed{\displaystyle P\left(\frac{1}{x}\right) \geq \frac{1}{P(x)}}$

Jake Lai - 5 years, 7 months ago

Very nice solution. Cauchy-Schwarz was the best method (in my opinion). +1

Sharky Kesa - 5 years, 7 months ago

Why thank you! I'm not entirely adept at utilising Cauchy-Schwarz to its fullest extent but I'm quite happy with the results.

Is this problem an original one? If so, well-posed!

Jake Lai - 5 years, 7 months ago

@Jake Lai – Yes, it was an original problem but no doubt other people have seen this before. I just used C-S to make this problem.

Sharky Kesa - 5 years, 7 months ago

Great Solution!

Ingenious use of the C-S inequality!

Sualeh Asif - 5 years, 7 months ago

Nice Solution! I like the way you used Cauchy-Schwarz inequality to prove this.

Surya Prakash - 5 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$