Show that $\int_{1}^\infty \dotsb\int_{1}^\infty \frac{dx_1 \dotsb dx_n}{x_1^{\alpha_1}+\dotsb + x_n^{\alpha_n}}<\infty$ Here's my solution to an old qualifier problem. Would you tackle it differently? Is there a flaw in my work?
Suppose that $\alpha_1, \dotsc, \alpha_n$ are positive numbers such that $$\frac1{\alpha_1}+ \dotsb + \frac1{\alpha_n}<1.$$
Show that $$\int_{1}^\infty \dotsb\int_{1}^\infty \frac{dx_1 \dotsb dx_n}{x_1^{\alpha_1}+\dotsb + x_n^{\alpha_n}}<\infty.$$
My answer: $x_1^{\alpha_1}+\dotsb + x_n^{\alpha_n} \overset{\text{(AM $\geq$ GM)}}{\geq} n\cdot \sqrt[n]{x_1^{\alpha_1}\dotsb x_n^{\alpha_n}}$
$$\Rightarrow \int_{1}^\infty \dotsb\int_{1}^\infty \frac{dx_1 \dotsb dx_n}{x_1^{\alpha_1}+\dotsb + x_n^{\alpha_n}} \leq \frac1n \int_{1}^\infty \dotsb\int_{1}^\infty \frac{dx_1 \dotsb dx_n}{x_1^{\alpha_1/n}\dotsb  x_n^{\alpha_n/n}},$$ 
at which point if we switch to hyperspherical coordinates
$$\leq \frac1n \int_{S^{n-1}} \int_{1}^\infty \frac{dx_1 \dotsb dx_n}{r^{\frac{\alpha_1+\dotsb + \alpha_n}{n}}|f(\sigma)|}r^{n-1} dr\,d\sigma,\tag{1}$$
where $\sigma$ is the position on the unit $(n-1)$ sphere in $\mathbb{R}^n$ and $|f(\sigma)|\leq1$. Note I have chosen a crude bound by integrating over the entire area outside the unit ball. Now 
$$\frac1{\alpha_1}+ \dotsb + \frac1{\alpha_n}<1 \\\Rightarrow \frac{n}{\frac1{\alpha_1}+ \dotsb + \frac1{\alpha_n}}>n\\\overset{\text{AM $\geq$ HM}}{\Rightarrow}\frac{\alpha_1 + \dotsb+\alpha_n}{n}>n$$ 
so the integral (1) is integrating a power of $r$ which is $>1$ in the denominator, so it converges.
 A: IMHO the main flaw in your logic is that you have generated a sufficient but not a necessary condition for the original integral to converge.
You write:
$$
  \int_1^\infty \cdots \int_1^\infty
  \frac{dx_1 \cdots dx_n}{x_1^{\alpha_1}+\cdots + x_n^{\alpha_n}}
  \leq \frac{1}{n} \int_{1}^\infty \cdots \int_{1}^\infty
  \frac{dx_1 \cdots dx_n}{x_1^{\alpha_1/n}\cdots x_n^{\alpha_n/n}}
$$
But:
$$
\frac{1}{n} \int_{1}^\infty \cdots \int_{1}^\infty
\frac{dx_1 \cdots dx_n}{x_1^{\alpha_1/n}\cdots x_n^{\alpha_n/n}} =
\frac{1}{n} \int_{1}^\infty \frac{dx_1}{x_1^{\alpha_1/n}} \cdots \int_{1}^\infty \frac{dx_n}{x_n^{\alpha_n/n}}
$$
And:
$$
\int_{1}^\infty \frac{dx_i}{x_i^{\alpha_i/n}}  = \left[ \frac{x_i^{1-\alpha_i/n}}{1-\alpha_i/n} \right]_1^\infty =
 \left\{ \begin{array}{ll} \frac{1}{1-\alpha_i/n} & \mbox{if} \quad \alpha_i > n \\
 \infty & \mbox{if} \quad \alpha_i \le n \end{array} \right.
$$
So instead of $1/\alpha_1 + \cdots + 1/\alpha_n < 1$ we have
$(1/\alpha_1 < 1/n) \wedge \cdots \wedge (1/\alpha_n < 1/n)$, which is a much stronger condition.
A: As already said on the page, the AM-GM inequality yields a sufficient condition too strong to be necessary. Instead one can rely directly on the change of variable $x_k^{\alpha_k}=ru_k^2$ with $r$ in $(1,\infty)$ and $u=(u_k)$ in the unit sphere $\sum\limits_ku_k^2=1$. Then the Jacobian of the transformation $(x_k)\to(r,u)$ is complicated to write down completely but it has the form $\prod\limits_k\mathrm dx_k=r^{\beta-1}\mathrm dr\mathrm d\mu(u)$ with $\beta=\sum\limits_k\frac1{\alpha_k}$, where $\mu$ is some finite measure on the unit sphere. Then $\sum\limits_kx_k^{\alpha_k}=r$ hence the integral to be considered converges if and only if
$$
\int^\infty\frac{r^{\beta-1}\mathrm dr}r
$$
converges, that is, if and only if $\beta\lt1$. 
A: The AM-GM could be modified as follows.
Denote
$$p:=\left(\frac{1}{\alpha_1}+\cdots+\frac{1}{\alpha_n}\right)^{-1}>1,$$
and denote 
$$y_i=\alpha_i\log x_i,\quad\lambda_i=\frac{p}{\alpha_i}\in (0,1), \quad i=1,\cdots,n.$$
Since $\sum_{i=1}^n\lambda_i=1$, by Jensen's inequality for the exponential function,
$$\sum_{i=1}^n x_i^{\alpha_i}\ge \sum_{i=1}^n \lambda_i e^{y_i}\ge \exp\left(\sum_{i=1}^n \lambda_i y_i\right)=\prod_{i=1}^n x_i^p.\tag{1}$$
By $(1)$ and Fubini's theorem, the original integral is bounded by 
$$\left(\int_1^\infty\frac{dx}{x^p}\right)^n=\frac{1}{(p-1)^n}.$$
A: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{{\rm e}^{#1}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\sgn}{{\rm sgn}}%
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert #1 \right\vert}$
$\ds{%
{\cal J}
\equiv
\int_{1}^{\infty}\cdots\int_{1}^{\infty}
{\dd x_{1}\ldots\dd x_{n}
 \over
 {x_{1}^{\alpha_{1}} + \cdots + x_{n}^{\alpha_{n}}}}\ <\ \infty:\ {\large ?}}$

\begin{align}
{\cal J}
&=
\int_{1}^{\infty}\cdots\int_{1}^{\infty}\bracks{%
\int_{0}^{\infty}\expo{-\pars{x_{1}^{\alpha_{1}} + \cdots + x_{n}^{\alpha_{n}}}\mu}\,\dd\mu}
\dd x_{1}\ldots\dd x_{n}
\\[3mm]&=
\int_{0}^{\infty}\dd\mu\prod_{i = 1}^{n}\int_{1}^{\infty}
\expo{-\pars{x^{\alpha_{i}}}\mu}\,\,\dd x
=
\int_{0}^{\infty}\dd\mu\prod_{i = 1}^{n}
{1 \over \alpha_{i}\,\mu^{1/\alpha_{i}}}
\int_{\mu}^{\infty}\expo{-x}\,x^{\pars{1/\alpha_{i}} - 1}\,\,\dd x
\\[3mm]&=
\int_{0}^{\infty}\prod_{i = 1}^{n}
{\Gamma\pars{1/\alpha_{i}\,,\,\mu} \over \alpha_{i}\,\mu^{1/\alpha_{i}}}
\,\dd\mu
=
\int_{0}^{\infty}
{1 \over \mu^{1/\alpha_{1} + \cdots + 1/\alpha_{n}}}
\prod_{i = 1}^{n}
{\Gamma\pars{1/\alpha_{i}\,,\,\mu} \over \alpha_{i}\,}
\,\dd\mu
\end{align}

$\Gamma\pars{z,a}$ is the
$\it\mbox{Incomplete Gamma function}\ \pars{~\Re\, z > 0~}$ and
$\Gamma\pars{z} \equiv \Gamma\pars{z,0}$ is the $\it\mbox{Gamma function}$. 


*

*$\large\mu\ \gtrsim\ 0$:
$$
{1 \over \mu^{1/\alpha_{1} + \cdots + 1/\alpha_{n}}}
\prod_{i = 1}^{n}
{\Gamma\pars{1/\alpha_{i}\,,\,\mu} \over \alpha_{i}\,}\ 
\sim\
{1 \over \mu^{\Lambda}}
\prod_{i = 1}^{n}
{\Gamma\pars{1/\alpha_{i}} \over \alpha_{i}\,}\,,
\quad
\Lambda \equiv {1 \over \alpha_{1}} + \cdots + {1 \over \alpha_{n}} < 1
$$
Since $0 < \Lambda < 1$, the integral ~ $\mu^{1 - \Lambda}$ when $\mu \gtrsim 0$


*$\large \mu\ \gg\ 1:$

In this case,
$\ds{%
\Gamma\pars{{1 \over \alpha_{i}},\mu}
\sim
\mu^{\pars{1/\alpha_{i}} - 1}\,\expo{-\mu}}$

$$
{1 \over \mu^{1/\alpha_{1} + \cdots + 1/\alpha_{n}}}
\prod_{i = 1}^{n}
{\Gamma\pars{1/\alpha_{i}\,,\,\mu} \over \alpha_{i}\,}\ 
\sim\
{\expo{-n\mu} \over \mu^{n}}
\prod_{i = 1}^{n}{1 \over \alpha_{i}\,}\,,
$$


So far, we can conclude that the condition
$\ds{\sum_{i = 1}{1 \over \alpha_{i}} < 1}$ guarantees the convergence of the integral.
