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# Not enough information at all!

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 $$x > 0$$.

Note by Sharky Kesa
1 year, 2 months ago

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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:

$\boxed{\displaystyle P\left(\frac{1}{x}\right) \geq \frac{1}{P(x)}}$ · 1 year, 2 months ago

Very nice solution. Cauchy-Schwarz was the best method (in my opinion). +1 · 1 year, 2 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! · 1 year, 2 months ago

Yes, it was an original problem but no doubt other people have seen this before. I just used C-S to make this problem. · 1 year, 2 months ago

Great Solution!

Ingenious use of the C-S inequality! · 1 year, 2 months ago

Nice Solution! I like the way you used Cauchy-Schwarz inequality to prove this. · 1 year, 2 months ago

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