# Finding a minimum

I would like to minimize the function $f(x_1,\dots,x_k) = x_1 + x_2 / x_1 + \dots + x_k / x_{k-1} + M / x_k$ where $M>0$ and $x_k \ge \dots \ge x_1 \ge 1$.

First I looked at a simpler function $g(x_1,x_2) = x_1 + M/x_2$. Computing the partial derivatives for $x_1$ and $x_2$ and finding the respective roots shows that the minimum value is attained when $x_1 = x_2 = \sqrt{M}$.

However, for the general case when $k$ is larger and in particular for $k=\log M$, it seems that setting $x_i = 2^{i-1}$ yields $2\log M + 1$ as the value of $f$, which I also suspect to be the global minimum.

If so, how can I formally show this?

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Hint: try using AM-GM inequality, which gives minimum when equality is there. –  Macavity Apr 5 '13 at 14:35

$x_1 + x_2 / x_1 + \dots + x_k / x_{k-1} + M / x_k\ge (k+1)(x_1.\frac{x_2}{x_1}.\frac{x_3}{x_2}\dots\frac{x_k}{x_{k-1}}\frac{M}{x_k})^{1/(k+1)}=(k+1)M^{1/(k+1)}$

This is by using A.M.-G.M.

Equality holds when $x_1=\frac{x_2}{x_1}=\frac{x_3}{x_2}=\dots=\frac{x_k}{x_{k-1}}=\frac{M}{x_k}$

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By AM-GM, we have: $$\frac{x_1 + \frac{x_2}{x_1} + \dots + \frac{x_k}{x_{k-1}} + \frac{M}{x_k}}{k+1} \ge \left( x_1 \cdot \frac{x_2}{x_1} \dots \frac{x_k}{x_{k-1}} \cdot \frac{M}{x_k}\right)^{\frac{1}{k+1}} = M^{\frac{1}{k+1}}$$
Equality is the minimum we can hope for, so the minimum is $(k+1)M^{\frac{1}{k+1}}$, when $$x_1 = \frac{x_2}{x_1} = \dots = \frac{x_k}{x_{k-1}} = \frac{M}{x_k}$$ (This gives a G.P. for $x_i$, with common ratio $x_1$. So $x_i = x_1^i$ and $M = x_1^{k+1}.$)