$a_1,a_2,...,a_n$ 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 $a_1a_2a_3...a_n =1$
 A: \begin{align}
&a_1 + \sqrt{a_2} + \sqrt[3]{a_3} + ... + \sqrt[n]{a_n} \\
=& \sum_{k=1}^n \sum_{j=1}^k \frac{1}{k} \sqrt[k]{a_k}\\
\ge & \frac{n(n+1)}{2} \left(\prod_{k=1}^n(\frac{1}{k}\sqrt[k]{a_k})^k\right)^{\frac{2}{n(n+1)}} \\
= & \frac{n(n+1)}{2} \left(\prod_{k=1}^n(\frac{1}{k^k}{a_k})\right)^{\frac{2}{n(n+1)}} \\
= & \frac{n(n+1)}{2} \left(\prod_{k=1}^n\frac{1}{k^k}\right)^{\frac{2}{n(n+1)}} \\
\ge & \frac{n(n+1)}{2} \left(\prod_{k=1}^n\frac{1}{n^k}\right)^{\frac{2}{n(n+1)}} \\
= & \frac{n(n+1)}{2} \left(\frac{1}{n^{\frac{n(n+1)}{2}}}\right)^{\frac{2}{n(n+1)}} \\
= &\frac{n+1}{2}
\end{align}
A: 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 $$\frac{n+1}{2}\leq\sum_{i=1}^{n}\frac{i}{n}{a_{i}}^{1/i}\leq\sum_{i=1}^{n}{a_{i}}^{1/i}.
 $$
A: 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
$$
\sum_{k=1}^na_k^{1/k}\ge\frac{n+1}2
$$
