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As part of a research problem I am working on, I need to show the following inequality. Let $x=(x_1,\dots,x_K)$ with $x_i > 0$ for all $i$. Then, I wish to show that $$ \frac{1}{K^2}\sum_{i=1}^K x_i \geq K \left(\sum_{i=1}^K x_i^{-1/2}\right)^{-2}.$$ For $K=2$ this is easy to show with simple algebra, but for the general case I haven't managed to find a proof. Note that this inequality can be rewritten as $$ \sum_{i,j,k}^K \frac{x_k}{\sqrt{x_ix_j}} \geq K^3. $$ Numerically it always seems to clearly hold. Using the fact that the $1/2$-norm is higher than the $1$-norm, I could show that it is higher than $K^2$, but not any $K^3$.

Any help would be greatly appreciated!

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Your sum $S$ also equals $$\sum_{i,j,k}\frac{x_j}{\sqrt{x_ix_k}}\qquad\text{and}\qquad\sum_{i,j,k}\frac{x_i}{\sqrt{x_jx_k}}.$$ Therefore $$3S=\sum_{i,j,k}\left(\frac{x_i}{\sqrt{x_jx_k}}+\frac{x_j}{\sqrt{x_ix_k}}+\frac{x_k}{\sqrt{x_ix_j}}\right)\ge 3K^2$$ on applying AM/GM to each summand.

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The inequality is equivalent to $$ \left(\frac{1}{K}\sum_{i=1}^K x_i^{-1/2}\right)^{-2} \le \frac{1}{K}\sum_{i=1}^K x_i $$ and that is a special case of the generalized mean inequality: $$ M_p(x_1, \ldots, x_K) \le M_q (x_1, \ldots, x_K) $$ for $p = -\frac 12 < q = 1$.

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Because by Holder: $$\left(\sum_{i=1}^K\frac{1}{\sqrt{x_i}}\right)^2\sum_{i=1}^Kx_i\geq\left(\sum_{i=1}^K\sqrt[3]{\left(\frac{1}{\sqrt{x_i}}\right)^2x_i}\right)^3=K^3.$$

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