Prove that $\int \limits_{0}^{\infty} x^p e^{-g(x)/x} dx \leq e^{p+1} \int \limits_{0}^{\infty} x^p e^{-g'(x)}dx$. 
For a convex function $g(0)=0$ and for any $-1<p<\infty$, prove that  $$\int \limits_{0}^{\infty} x^p e^{-g(x)/x} dx \leq e^{p+1} \int \limits_{0}^{\infty} x^p e^{-g'(x)}dx.\,\,\,\,\,\,(♣)$$

It the generalization of the not-so-famous Carleman's Integral Inequality, which states that :
$$\int \limits_0^\infty \text{exp}\left\{\frac{1}{x}\int \limits_{0}^{x }\ln(f(t))\,dt\right\}dx \leq e \int \limits_0 ^{\infty}f(x)\,dx.\,\,\,\,\,(♥)$$
$(♥) $ is a special case of $(♣) $ with $p=0$.
$(♥)$ is "OK-ayish" but $(♣)$ is very tricky to prove, in which I horribly fail. I can not find any suitable reference or method or something which can help me prove this monster.
Any help will be appreciated.
Thanks in Advance !  :-)

A proof can be found here, at section 2, but I'll go over how it works here. First, the convexity condition on $g$ is used, in particular fact 5 here to show that 
$$g(k x) \geq g(x) + (k-1) x g'(x)$$
for any $k>1$. Then, consider the integral
$$J = \int_0^A x^p \exp\left(-\frac{g(kx)}{kx}\right) dx$$ for $A>0$.
By a substitution, you can show 
$$\begin{align}J &= k^{-p-1}\int_0^{Ak} x^p \exp\left(-\frac{g(x)}{x}\right) dx \\
&\geq k^{-p-1}\int_0^{A} x^p \exp\left(-\frac{g(x)}{x}\right) dx\end{align}$$
On the other hand, use the convexity inequality on $g$ to show
$$J \leq \int_0^A x^p \exp\left(-\frac{g(x)}{kx} - \frac{(k-1)  g'(x)}{k}\right) dx$$
from which you can use Holder's inequality here in integral form to get
$$J \leq \left(\int_0^A x^p \exp\left(-\frac{g(x)}{x}\right) dx \right)^{1/k} \left(\int_0^A x^p \exp\left(-g'(x)\right) dx\right)^{(k-1)/k}$$
which should start to look familiar.
Putting our bounds on $J$ together, we get
$$k^{-p-1}\left(\int_0^A x^p \exp\left(-\frac{g(x)}{x}\right) dx \right)^{(k-1)/k} \leq  \left(\int_0^A x^p \exp\left(-g'(x)\right) dx\right)^{(k-1)/k}$$
So take the limit as $A \to \infty$ and rearrange to get
$$\int_0^\infty x^p \exp\left(-\frac{g(x)}{x}\right) dx \leq \left(k^{\frac k{k-1}}\right)^{p+1} \int_0^\infty x^p \exp\left(-g'(x)\right) dx$$
and taking the $k\to 1$ limit finishes off the answer!
