Show that this integral converges, Show that 
$$\int_1^{\infty}\int_1^{\infty}\cdots\int_1^{\infty} \frac{{\rm d}x_1\cdots {\rm d}x_n}{x_1^{\alpha_1} + \ldots + x_n^{\alpha_n}} < \infty$$
if $\frac{1}{\alpha_1} + \ldots + \frac{1}{\alpha_n} < 1 $
I'm guessing that, to make use of the inequality assumption given in the problem statement, I should look to make a change of variables.  
Any ideas are welcome. Thanks!
 A: Start with a change of variables 
$$y_i = x_i^{\frac{\alpha_i}{2}}\implies {\rm d}x_i = \frac{2{\rm d}y_i}{\alpha_i y_i^{1- \frac{2}{\alpha_i}}}$$
to get the integral
$$\frac{2^{n}}{\alpha_1\cdots \alpha_n}\int_0^1\cdots \int_0^1 \frac{1}{y_1^2 + \ldots + y_n^2}\prod_{i=1}^n\frac{{\rm d}y_i}{y_i^{1- \frac{2}{\alpha_i}}}$$
The integral above is bounded by the integral of the same integrand over the spherically symmetric region $r^2 \equiv y_1^2 + \ldots + y_n^2 \geq 1$ (since the integrand is stricktly positive and contains $[1,\infty)^n$) so let's focus on this integral as this makes the integration limits simple when changing to spherical coordinates.
Since we are only interested in establishing convergence we are going to change to spherical coordinates $\{r,\theta_1,\ldots,\theta_{n-1}\}$ and only focus on the radial ($r$) integral as this is the only possible source of divergence since the integrand has no sigularities in the integration region.
The relevant transformation formulas to spherical coordinates can be found in this answer:
$$\prod_{i=1}^n {\rm d}y_i = r^{n-1} {\rm d}r\prod_{i=1}^{n-1}\sin^{n-1-i}(\theta_i){\rm d}\theta_i$$ 
$$\prod_{i=1}^n y_i^{1-\frac{2}{\alpha_i}} = r^{n - \sum_{i=1}^n\frac{2}{\alpha_i}}\prod_{i=1}^{n-1}\cos^{1-\frac{2}{\alpha_i}}(\theta_i)\sin^{n-i-\sum_{j=i+1}^n\frac{2}{\alpha_i}}(\theta_i)$$ 
From this it follows that the radial integral becomes
$$\int_1^\infty \frac{1}{r^2}\cdot r^{n-1}\cdot \frac{1}{r^{n-\sum_{i=1}^n\frac{2}{\alpha_i}}}{\rm d}r =  \int_1^\infty \frac{{\rm d}r}{r^{3-2\sum_{i=1}^n\frac{1}{\alpha_i}}}$$
which converges if $3-2\sum_{i=1}^n\frac{1}{\alpha_i} > 1 \implies \sum_{i=1}^n\frac{1}{\alpha_i} < 1$.
A: We first show that for any $\alpha > 1$, there exists $C(\alpha) > 0$ such that 
$$\int_1^\infty \frac{dx}{a + x^\alpha} \leq C(\alpha) a^{\frac1{\alpha}-1},$$
for any $a >1$. Indeed, by a change of variable $x = a^{\frac1\alpha} t$, we have
$$\int_1^\infty \frac{dx}{a + x^\alpha} = a^{\frac1\alpha -1}\int_{a^{-\frac1\alpha}}^\infty \frac{dt}{1 + t^{\alpha}} \leq a^{\frac1\alpha -1}\int_{0}^\infty \frac{dt}{1 + t^{\alpha}},$$
hence we can take $C(\alpha) = \int_{0}^\infty \frac{dt}{1 + t^{\alpha}}.$
Since $x_i \geq 1$ for $i =1,\ldots,n$ then we have
$$\int_1^\infty \frac{d x_n}{x_1^{\alpha_1} + x_2^{\alpha_2}+\ldots +x_n^{\alpha_n}} \leq C(\alpha_n) (x_1^{\alpha_1} +\ldots +x_{n-1}^{\alpha_{n-1}})^{\frac{1-\alpha_n}{\alpha_n}}.$$
Denote $\beta_i = \alpha_i (\alpha_n -1)/\alpha_n$, then $\beta_i > 0$ and 
$$\frac1{\beta_1} + \ldots + \frac1{\beta_{n-1}} < 1.$$
It is easy to see that 
$$(x_1^{\alpha_1} +\ldots +x_{n-1}^{\alpha_{n-1}})^{\frac{\alpha_n-1}{\alpha_n}}\geq \frac1{n-1} (x_1^{\beta_1} + \ldots (x_{n-1}^{\alpha_{n-1}}).$$
Hence
$$\int_1^\infty\cdots\int_1^\infty \frac{dx_1\ldots dx_n}{x_1^{\alpha_1} + x_2^{\alpha_2}+\ldots +x_n^{\alpha_n}}\leq C(\alpha_n)(n-1)\int_1^\infty\cdots\int_1^\infty \frac{dx_1\ldots dx_{n-1}}{x_1^{\beta_1} + \ldots +x_{n-1}^{\beta_{n-1}}}.$$
Repeating the arguments above, we see that this integral converges.
