2. Zadatak

Uz oznaku $x_{n+1}=x_1$ imamo:

$$\begin{aligned}\sum_{i=1}^n\frac{x_i^2}{x_i+x_{i+1}}&=\sum_{i=1}^n\frac{x_i^2+x_ix_{i+1}-x_ix_{i+1}}{x_i+x_{i+1}}\\&=\sum_{i=1}^n\left(x_i-\frac{x_ix_{i+1}}{x_i+x_{i+1}}\right)\ge\sum_{i=1}^n x_i-\sum_{i=1}^n\frac{\left(\frac{x_i+x_{i+1}}2\right)^2}{x_i+x_{i+1}}\\&=1-\frac14\sum_{i=1}^n(x_i+x_{i+1})=\frac12.\end{aligned}$$

Drugo rješenje. Uz oznaku $x_{n+1}=x_1$ dobivamo:

$$\begin{aligned}\sum_{i=1}^n\frac{x_i^2}{x_i+x_{i+1}}&=\sum_{i=1}^n\frac{x_i^2-x_{i+1}^2+x_{i+1}^2}{x_i+x_{i+1}}\\&=\sum_{i=1}^n(x_i-x_{i+1})+\sum_{i=1}^n\frac{x_{i+1}^2}{x_i+x_{i+1}}\\&=\sum_{i=1}^n\frac{x_{i+1}^2}{x_i+x_{i+1}}.\end{aligned}$$

Odavde dobivamo

$$\sum_{i=1}^n\frac{x_i^2}{x_i+x_{i+1}}=\frac12\sum_{i=1}^n\frac{x_i^2+x_{i+1}^2}{x_i+x_{i+1}}\ge\frac14\sum_{i=1}^n(x_i+x_{i+1})=\frac12,$$

pri čemu smo koristili nejednakost između aritmetičke i kvadratne sredine.