# $a_1,a_2,…,a_n$ are positive real numbers, their product is equal to $1$, show: $\sum_{i=1}^n a_i^{\frac 1 i} \geq \frac{n+1}2$

it says to use the weighted AM-GM to solve it, because the inequality is not homogenous I've tried to use $$\lambda _ i = \frac{a_i^{\frac1i -1}}{\sum_{k=1}^n a_k^{\frac1k -1}}$$

this $\lambda$ is from the inequality:

$$\sum_{i=1}^n \lambda_i a_i \geq \Pi_{i=1}^n a_i^{\lambda_i}$$

the sum of all $\lambda_i$ must be 1. It didn't work and I'm stuck

To be more clear, I need to prove that

$$a_1 + \sqrt{a_2} + \sqrt[3]{a_3} + ... + \sqrt[n]{a_n} \geq \frac{n+1}{2}$$

and we know that $a_1a_2a_3...a_n =1$

Using the weighted AM-GM and recalling that $$\sum_{i=1}^{n}i=\frac{n\left(n+1\right)}{2}$$ we have $$\sum_{i=1}^{n}i{a_{i}}^{1/i}\geq\frac{n\left(n+1\right)}{2}\left(a_{1}a_{2}\cdots a_{n}\right)^{2/\left(n\left(n+1\right)\right)}=\frac{n\left(n+1\right)}{2}$$ hence $$\frac{n+1}{2}\leq\sum_{i=1}^{n}\frac{i}{n}{a_{i}}^{1/i}\leq\sum_{i=1}^{n}{a_{i}}^{1/i}.$$
Since $$\sum_{k=1}^n\frac{2k}{n^2+n}=1$$ the weighted AM-GM says that with weights $\frac{2k}{n^2+n}$, \begin{align} \sum_{k=1}^n\frac{2k}{n^2+n}a_k^{1/k} &\ge\prod\left(a_k^{1/k}\right)^{\frac{2k}{n^2+n}}\\ &=1 \end{align} However, since $$\frac{2n}{n^2+n}\ge\frac{2k}{n^2+n}$$ we have $$\sum_{k=1}^n\frac{2n}{n^2+n}a_k^{1/k}\ge1$$ which is equivalent to $$\sum_{k=1}^na_k^{1/k}\ge\frac{n+1}2$$