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A Cauchy Schwarz Problem

Cauchy Schwarz 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^+}}$

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