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

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.

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

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