**Problem:Let be a polynomial with non-negative coefficients.Prove that if for ,then the same inequality holds for each .**

**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 ,then

now

(**Cauchy Schwarz’s Inequality**)

now it is given that for ,so

therefore,

## No comments, be the first one to comment !