Mathematical Olympiad: Tricky Inequality. 
Now we have to use the Weighted AM-GM Inequality. In order to do so we need to find $\lambda_1,\lambda_2,....\lambda_n>0$ such that $\lambda_1+\lambda_2,....+\lambda_n=1$. Then the inequality should take the usual form: $$x_1\lambda_1+x_2\lambda_2,....+x_n\lambda_n\geq x_1^{\lambda_1}x_2^{\lambda_2}.....x_n^{\lambda_n}.$$

The natural choice, to deal with the RHS, would be something of the form $x_i=a^{1/i}$ and $\lambda_i=i$. Since we want $\sum \lambda_i=1$, we normalize, and take instead $\lambda_i=i/(\sum_{i=1}^n i)=\frac{2}{n(n+1)}i$. Then
$$\prod(a_i^{1/i})^{\lambda_i}=(\prod a_i)^{2/n(n+1)}=1,$$
and thus
$$1=\prod(a_i^{1/i})^\lambda_i\leq\sum\lambda_i a_i^{1/i}\leq\lambda_n\sum a_i^{1/i}=\frac{2}{n+1}\sum a_i^{1/i}$$
so multiplying both sides by $\frac{n+1}{2}$ gives the desired inequality.
