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Problem:Let $P(x)$ be a polynomial with non-negative coefficients.Prove that if $P(x)\cdot P(\frac{1}{x})\ge1$ for $x=1$,then the same inequality holds for each $\mathbb{R^+} x$.

Discussion: Cauchy Schwarz’s Inequality: Suppose for real numbers $\ a_{i},b_{i}$, where $\ i\in{1,2,\dots,n}$ we can say that $$\{\sum_{i=1}^{n}a_{i}^2\}\{\sum_{i=1}^{n}b_{i}^2\}=\sum_{i=1}^{n}\{a_{i}b_{i}\}^2$$.

Titu’s Lemma: Let $\ a_{i},b_{i}\in{\mathbb{R}}$ and let $\ a_{i},b_{i}>0$ for $\ i\in{1,2,\dots,n}$

$$\sum_{i=1}^{n}\frac{a_{i}^2}{b_{i}}\ge\frac{\{\sum_{i=1}^{n}a_{i}\}^2}{\sum_{i=1}^{n}b_{i}}$$

Proof of Cauchy Schwarz’s Inequality: We can write $\sum_{i=1}{n}a_{i}^2=\sum_{i=1}^{n}\frac{a_{i}^2 b_{i}^2}{b_{i}^2}\ge\frac{\{\sum_{i=1}^{n}a_{i}b_{i}\}^2}{\sum_{i=1}^{n}b_{i}^2}$    (using Titu’s lemma)

$=>\{\sum_{i=1}^{n}a_{i}^2\}\{\sum_{i=1}^{n}b_{i}^2\}\ge \{\sum_{i=1}^{n}\{a_{i}b_{i}\}^2$

Solution: Let $P(x)=a_{n}x^n+a_{n-1}x^{n-1}+\dots+a_{1}x+a_{0}$,then $P(\frac{1}{x})=a_{n}\frac{1}{x^n}+a_{n-1}\frac{1}{x^{n-1}}+\dots+a_{1}\frac{1}{x}+a_{0}$

now $P(x)\cdot P(\frac{1}{x})=\{a_{n}x^n+a_{n-1}x^{n-1}+\dots+a_{1}x+a_{0}\}\{a_{n}\frac{1}{x^n}+a_{n-1}\frac{1}{x^{n-1}}+\dots+a_{1}\frac{1}{x}+a_{0}\}=\{\{\sqrt{a_{n}x^n}\}^2+\{\sqrt{a_{n-1}x^{n-1}}\}^2+\dots+\{\sqrt{a_{1}x}\}^2+\{\sqrt{a_{0}}\}^2\} \{\{\sqrt{a_{n}\frac{1}{x^n}}\}^2+\{\sqrt{a_{n-1}\frac{1}{x^{n-1}}}\}^2+\dots+\{\sqrt{a_{1}\frac{1}{x}}\}^2+\{\sqrt{a_{0}}\}^2\} \ge\{a_{n}+a_{n-1}+\dots+a_{1}+a_{0}\}^2$
(Cauchy Schwarz’s Inequality)

now it is given that $P(x)\cdot P(\frac{1}{x})\ge1$ for $x=1$,so $P(1)^2\ge 1=>\{a_{n}+a_{n-1}+\dots+a_{1}+a_{0}\}^2\ge1$

therefore,$P(x)\cdot P(\frac{1}{x})\ge1$ $\forall x\in{\mathbb{R^+}}$