0
$\begingroup$

I have a short question.

Is it possible to compute the following:

$ \inf \left\lbrace \frac{\sqrt{\sum_{i=1}^n x_i}}{\sum_{i=1}^n \sqrt{x_i}} \, \middle| \, \forall i: x_i>0 \right\rbrace $

An upper bound for this value is $ 1/\sqrt{n} $, but I'm not sure whether this is also the value of the infimum.

  • 0
    probably seem more familiar if you make new (positive) variables $y_i$ with $x_i = y_i^2$ – 2017-02-07
  • 0
    For $n=2$, I can see how you can prove that $1/\sqrt2$ is $\inf$, since your expression is decreasing in $x_{i}$ if $x_{i}x_{j}$ and $\lim_{x\rightarrow\infty}\frac{\sqrt{2x}}{2\sqrt x}=\frac{1}{\sqrt2}$. For general $n$? But observe for $n=3$ the sign of derivative w.r.t. $x_{i}$ depends on $x_{i}\lessgtr \frac{(x_{j}+x_{k})^2}{(\sqrt x_{j}+\sqrt x_{k})^{2}}$. – 2017-02-07

1 Answers 1

1

By the root-mean square inequality (a special case of the generalized AM-GM):

$$ \sqrt{\frac{\sum_{i=1}^n x_i}{n}} \;\ge\; \frac{\sum_{i=1}^n \sqrt{x_i}}{n} \quad \iff \quad \frac{\sqrt{\sum_{i=1}^n x_i}}{\sum_{i=1}^n \sqrt{x_i}} \ge \frac{1}{\sqrt{n}} $$

Equality is attained iff $\;x_i = x_j\;$ so $\;\cfrac{1}{\sqrt{n}}\;$ is not only an infimum, but an actual minimum as well.