The classical formulation of Young's inequality is \begin{equation*} xy \leq \frac{x^p}{p} + \frac{y^q}{q}, \quad \text{where} \quad \frac{1}{p} + \frac{1}{q} = 1. \end{equation*} It's fairly trivial to extend this to \begin{equation*} x^ay^b \leq \frac{a}{a+b} x^{a+b} + \frac{b}{a+b} y^{a+b}. \end{equation*} It should be possible to find constants where the following holds: \begin{equation*} \prod_{i=1}^n x_i^{a_i} \leq \sum_{i=1}^n C_i x_i^m, \end{equation*} where $\sum_{i=1}^n a_i = m$. I'm really only interested in the case where $n = 3$; however, a generalized result would be nice. Any help is much appreciated.
2026-03-25 09:38:36.1774431516
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Generalization of algebraic Young's inequality to n variables.
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You can show by induction that for all $x_i > 0, \, r_i> 1$ for which $\sum_{i}r_i^{-1} = 1$ the estimate holds $$ \prod_{j=1}^n x_i \le \sum_{i=1}^n \frac{x_i^{r_i}}{r_i} $$ Replace $x_i$ with $x_i^{a_i}$ and the desired family of estimates follows.
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If you write $x_i = y_i^m$ and $a_i = mb_i$, the inequality becomes the weighted AM–GM inequality $$ \prod_i y_i^{b_i} \leq \sum_i b_i y_i, $$ which is probably most easily proved by applying Jensen's inequality to $-\log{x}$.
Yes, it's possible for positives $x_i$ and $a_i$.
Indeed, let $C_i=\frac{a_i}{m}$.
Thus, $\sum\limits_{i=1}^nC_i=1$ and by AM-GM we obtain: $$\sum_{i=1}^nC_ix_i^m\geq\prod_{i=1}^n\left(x_i^m\right)^{C_i}=\prod_{i=1}^n\left(x_i^m\right)^{\frac{a_i}{m}}=\prod_{i=1}^nx_i^{a_i}$$ and we are done!