Problem 1428

Given $\because a_{i} \in \mathrm{R}^{+}(i=1,2, \cdots n)$, and $\sum_{i=1}^{n} a_{i}=1$,
$$\begin{aligned} \therefore \sum_{i=1}^{n} \frac{x_{i}^{2}}{a_{i}} & =\left(\sum_{i=1}^{n} a_{i}\right) \cdot\left(\sum_{i=1}^{n} \frac{x_{i}^{2}}{a_{i}}\right) \\ & =\left[\sum_{i=1}^{n}\left(\sqrt{a_{i}}\right)^{2}\right] \cdot\left[\sum_{i=1}^{n}\left(\frac{x_{i}}{\sqrt{a_{i}}}\right)^{2}\right] \\ & \geqslant\left[\sum_{i=1}^{n} \sqrt{a_{i}} \cdot \frac{x_{i}}{\sqrt{a_{i}}}\right]^{2} \\ & =\left(\sum_{i=1}^{n} x_{i}\right)^{2} \end{aligned}$$

Equality holds if and only if $\frac{\frac{x_{1}}{\sqrt{a_{1}}}}{\sqrt{a_{1}}}=\frac{\frac{x_{2}}{\sqrt{a_{2}}}}{\sqrt{a_{2}}}=\cdots=\frac{\frac{x_{n}}{\sqrt{a_{n}}}}{\sqrt{a_{n}}}$, i.e., $\frac{x_{1}}{a_{1}}=\frac{x_{2}}{a_{2}}=\cdots=\frac{x_{n}}{a_{n}}$.