# Dorin Marghidanu's Inequality with Radicals

### Solution 1

Denote $\displaystyle \sqrt[i_k]{x_k}=a_k.$ We need to show that $\displaystyle \sum_{k=1}^{n}a_k\gt\sqrt[\small{\displaystyle \sum_{k=1}^{n}i_k}]{\prod_{k=1}^{n}a_k^{i_k}}.$

By Jensen's inequality, $\displaystyle \ln\left(\frac{\displaystyle \sum_{k=1}^{n}i_ka_k}{\displaystyle \sum_{k=1}^{n}i_k}\right)\ge\ln\left(\prod_{k=1}^{n}a_k^{i_k}\right).$ Hence, all we have to do is prove that $\displaystyle \sum_{k=1}^{n}a_k\gt\frac{\displaystyle \sum_{k=1}^{n}i_ka_k}{\displaystyle \sum_{k=1}^{n}i_k},\,$ which is true since $\displaystyle \sum_{k=1}^{n}a_k\gt\max a_k\ge\frac{\displaystyle \sum_{k=1}^{n}i_ka_k}{\displaystyle \sum_{k=1}^{n}i_k}.$

### Solution 2

Suppose the opposite, i.e.,

$\displaystyle\sum_{k=1}^{n}\sqrt[i_k]{x_k} \le \sqrt[\small{\displaystyle \sum_{k=1}^{n}i_k}]{\prod_{k=1}^{n}x_k}.$

Then, for each $k,$ $\displaystyle \sqrt[i_k]{x_k} \lt \sqrt[\small{\displaystyle \sum_{k=1}^{n}i_k}]{\prod_{k=1}^{n}x_k}$ such that $\displaystyle x_k \lt \sqrt[\small{\displaystyle \sum_{k=1}^{n}i_k}]{\left(\prod_{k=1}^{n}x_k\right)^{i_k}},$ $k=\overline{1,n}.$

Multiplying, we get

$\displaystyle \prod_{k=1}^{n}x_k\lt\sqrt[\small{\displaystyle \sum_{k=1}^{n}i_k}]{\left(\prod_{k=1}^{n}x_k\right)^{i_1+i_2+\ldots+i_n}}=\prod_{k=1}^{n}x_k.$

### Solution 3

We'll use the Multinomial expansion:

$\displaystyle \left(\sum_{k=1}^{m}y_k\right)^{n}=\sum_{k_1+\ldots+k_m=n}\left(\frac{n!}{k_1!\cdots k_m!}\right)\prod_{t=1}^{m}y_t^{k_t},$

where $\displaystyle n=\sum_{k=1}^{m}i_k\,$ and $\displaystyle y_t=x_t^{\frac{1}{i_t}},\,$ $t=1,2,\ldots,m.\,$ With $m=n,$ we have

$\displaystyle \left(\sum_{k=1}^{n}\sqrt[i_k]{x_k}\right)^{\sum_{k=1}^{n}i_k}=\sum_{k_1+\ldots+k_n=\sum_{s=1}^{n}i_s}\left(\frac{\displaystyle \left(\sum_{k=1}^{n}i_k\right)!}{\displaystyle \prod_{s=1}^{n}k_s!}\right)\prod_{t=1}^{n}\left[x_t^{\frac{1}{i_t}}\right]^{k_t}\gt\prod_{k=1}^{n}x_k,$

because $\displaystyle \frac{\displaystyle n!}{\displaystyle \prod_{s=1}^{n}k_s!}=\frac{\displaystyle \left(\sum_{k=1}^{n}i_k\right)!}{\displaystyle \prod_{s=1}^{n}k_s!}\gt 1\,$ and $\displaystyle \prod_{t=1}^{n}\left[x_t^{\frac{1}{i_t}}\right]^{k_t}=\prod_{k=1}^{n}x_k,\,$ when $i_t=k_t.$

### Solution 4

\displaystyle \begin{align} LHS &\gt \sum_{k=1}^n\frac{i_k}{\sum_{j=1}^n i_j}\sqrt[i_k] x_k \\ &=\frac{1}{\sum_{j=1}^n i_j}\left(\sum_{k=1}^n i_k\sqrt[i_k]x_k\right) \\ &\geq\sqrt[\sum_{j=1}^n i_j]{\prod_{k=1}^n \left(\sqrt[i_k]x_k\right)^{i_k}}~\text{(AM-GM)} \\ &= RHS. \end{align}

### Acknowledgment

This problem was been kindly posted at the CutTheKnotMath facebook page by Dorin Marghidanu. Solution 1 is by Leo Giugiuc; Solution 2 is by Dorin Marghidanu; Solution 3 is by Marian Dinca; Solution 4 is by Amit Itagi and, independently, by Dorin Marghidanu.