Školjka

Prema $CSB$ nejednakosti za sve realne brojeve $b_1, b_2,\ldots, b_n$ vrijedi $$n\bigg(\sum_{i=1}^{n}b_i^2\bigg)=\underbrace{(1+1+\cdots+1)}_{\text{n puta}}(b_1^2+\cdots+b_n^2)\stackrel{CSB}{\ge}(b_1+b_2+\cdots+b_n)^2$$ Korjenovajući obje strane imamo $$b_1+b_2+\cdots+b_n\le\sqrt{n}\sqrt{b_1^2+\cdots+b_n^2}$$ Uzimajući $\forall k$ $$b_k=\frac{x_k}{1+x_1^2+\cdots+x_k^2}$$ Dobivamo $$\frac{x_1}{1+x_1^2} + \frac{x_2}{1+x_1^2 + x_2^2} + \cdots + \frac{x_n}{1 + x_1^2 + \cdots + x_n^2}\le\sqrt{n}\sqrt{\bigg(\frac{x_1}{1+x_1^2}\bigg)^2 + \bigg(\frac{x_2}{1+x_1^2 + x_2^2}\bigg)^2 + \cdots + \bigg(\frac{x_n}{1 + x_1^2 + \cdots + x_n^2}\bigg)^2}$$ Preostaje pokazati $\bigg(\frac{x_1}{1+x_1^2}\bigg)^2 + \bigg(\frac{x_2}{1+x_1^2 + x_2^2}\bigg)^2 + \cdots + \bigg(\frac{x_n}{1 + x_1^2 + \cdots + x_n^2}\bigg)^2<1$. Primjetimo da za $k=1$ imamo $\bigg(\frac{x_1}{1+x_1^2}\bigg)=\frac{x_1^2}{(1+x_1^2)(1+x_1^2)}\le\frac{x_1^2}{(1+x_1^2)}=1-\frac{1}{1+x_1^2}$ te $\forall k\ge 2$ imamo \begin{align*} \bigg(\frac{x_k}{1 + x_1^2 + \cdots + x_k^2}\bigg)^2&=\frac{x_k^2}{(1 + x_1^2 + \cdots + x_k^2)^2}\\ &\le\frac{x_k^2}{(1 + x_1^2 + \cdots + x_k^2)(1 + x_1^2 + \cdots + x_{k-1}^2)}\\ &=\frac{(1 + x_1^2 + \cdots + x_k^2)-(1 + x_1^2 + \cdots + x_{k-1}^2)}{(1 + x_1^2 + \cdots + x_k^2)(1 + x_1^2 + \cdots + x_{k-1}^2)}\\ &=\frac{1}{1 + x_1^2 + \cdots + x_{k-1}^2}-\frac{1}{1 + x_1^2 + \cdots + x_k^2} \end{align*} Pa teleskopskom sumom dobivamo $$\sum_{k=1}^{n}\bigg(\frac{x_k}{1+x_1^2+\cdots+x_k^2}\bigg)^2\le1-\frac{x_n}{1 + x_1^2 + \cdots + x_n^2}<1$$ I to je to. Najsss!