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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 $a_1a_2a_3...a_n =1$

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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 $$\frac{n+1}{2}\leq\sum_{i=1}^{n}\frac{i}{n}{a_{i}}^{1/i}\leq\sum_{i=1}^{n}{a_{i}}^{1/i}. $$

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\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}

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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 $$ \sum_{k=1}^na_k^{1/k}\ge\frac{n+1}2 $$

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