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 \(x > 0\).

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 \(x > 0\).

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TopNewestLet \(n = \deg P\) 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 · 1 year ago

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– Sharky Kesa · 1 year ago

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

Is this problem an original one? If so, well-posed! – Jake Lai · 1 year ago

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– Sharky Kesa · 1 year 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.Log in to reply

Ingenious use of the C-S inequality! – Sualeh Asif · 1 year ago

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– Surya Prakash · 1 year ago

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