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Given that $b_1+b_2+\dots+b_n = 1$, how do I find the minimum value of $$\frac{x_1+x_2+\dots+x_n}{x_1^{b_1}x_2^{b_2}\dots x_n^{b_n}}?$$

For $n=2$ I used calculus and found the answer to be $$\frac{1}{b_1^{b_1}b_2^{b_2}}.$$ Extending the concept to higher values of $n$, the desired answer may be guessed as $$\prod\limits_{i=1}^n \left(\frac{1}{b_i}\right)^{b_i}.$$

Is there a better approach?

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You can look at how people have done the formatting by right clicking on the $\LaTeX$ and choosing "Show Source". Then put that in dollar signs. For example, joriki has just the expression you want for your guessed solution. – Ross Millikan Apr 21 '11 at 13:54
up vote 5 down vote accepted

I assume all $x_i$ and $b_i$ are positive. Put${x_i\over b_i}=:\lambda_i > 0$ and $\prod_i b_i^{b_i}=:B$. Then by the AGM inequality one has $$\prod_i x_i^{b_i}=B\ \prod_i \lambda_i^{b_i} \leq B\ \sum_i b_i\lambda_i = B\ \sum_i x_i$$ with equal sign iff all $\lambda_i$ are equal. This immediately leads to the conjectured result.

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I assume that the $x_i$ and $b_i$ are restricted to positive values? Also I'm assuming that you meant to write $\prod(1/b_i)^{b_i}$ for the guess.

You can find the result you guessed in much the same way that you found it for $n=2$. Setting the derivative with respect to $x_i$ to zero yields

$$x_i=b_i\sum_j x_j\;,$$

which is a homogeneous system of $n$ linear equations for the $n$ unknowns. Due to the condition $\sum b_i=1$, the rank is $n-1$, so there's a one-dimensional subspace of solutions, namely $x_i=\lambda b_i$ with arbitrary $\lambda$. The parameter $\lambda$ drops out of the function to be minimized, so you can take $\lambda=1$, i.e. $x_i=b_i$, which yields the minimum value you guessed.

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You only have proven that the points $x=\lambda b$ are somehow special. For what it's worth the inequality might go exactly the other way around. – Christian Blatter Apr 21 '11 at 9:19
@Christian: Well, the question says "how to find the minimum value" -- the definite article implies that it's already known that there is a minimum value :-) And of course if there is a minimum value, all derivatives must vanish, so this is the only candidate. – joriki Apr 21 '11 at 12:59

Since the $b_i$'s are fractional, I assume that $b_i,x_i\ge0$.

Due to the concavity of $\log(x)$, Jensen's Inequality says $$ \sum_{i=1}^nb_i\log(x_i)\le\log\left(\sum_{i=1}^nb_ix_i\right)\tag{1} $$ $(1)$ is true for any $x_i$, so we can substitute $x_i\mapsto\frac{x_i}{b_i}$: $$ \sum_{i=1}^nb_i\log(x_i/b_i)\le\log\left(\sum_{i=1}^nx_i\right)\tag{2} $$ Apply $e^x$ to both sides, $$ \prod_{i=1}^n\frac{x_i^{b_i}}{b_i^{b_i}}\le\sum_{i=1}^nx_i\tag{3} $$ Dividing both sides by $\prod\limits_{i=1}^nx_i^{b_i}$ yields $$ \prod_{i=1}^nb_i^{-b_i}\le\sum_{i=1}^nx_i\prod_{i=1}^nx_i^{-b_i}\tag{4} $$ By evaluating at $x_i=b_i$, we get that $(4)$ is sharp. That is, the minimum of the right hand side is the left hand side.

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