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Let $P$ be a polynomial with positive real coefficients. Prove that if $$ P\left( \frac{1}{x} \right) \geq \frac{1}{P(x)} $$ holds for $x = 1$, then it holds for every $x > 0$.

What I did:

I was thinking that it might be possible to use an inequality such as the AM-GM inequality but I'm not sure how. Any help would be appreciated.

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  • $\begingroup$ If what, then it holds . . . what, pray tell, is what? $\endgroup$ – Robert Lewis Feb 25 at 5:21
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Hint:

So $p(1)\geq 1$. Now by the Cauchy–Schwarz inequality the statement follows. Just write the polynomial in standard form: $$p(x) = a_nx^n+...+a_2x^2+a_1x+a_0$$

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Let $P(x)=a_0x^n+a_1x^{n-1}+...+a_n,$ where all $a_i>0$.

Thus, the condition gives $$P(1)\geq \frac{1}{P(1)}$$ or $$P(1)^2\geq1.$$ Id est, by C-S for all $x>0$ we obtain: $$P\left(\frac{1}{x}\right)P(x)=\left(\frac{a_0}{x^n}+\frac{a_1}{x^{n-1}}+...+a_n\right)(a_0x^n+a_1x^{n-1}+...+a_n)\geq$$ $$\geq(a_0+a_1+...+a_n)^2=P(1)^2\geq1$$ and we are done!

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  • $\begingroup$ Is the "≥" a duplicate in the part at the end? Or did you leave something out? $\endgroup$ – user634856 Feb 10 at 1:48

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