An inequality by using general Hölder's inequality Let $w$ be a weight function and $\frac{1}{p}=\frac{1}{p_1}+\frac{1}{p_2}+\dots+\frac{1}{p_n}$. For suitable functions let $\overrightarrow{f}=(f_1,f_2,\dots,f_n)$ and operator $T$ satisifies the condition 
$$
T(\overrightarrow{f})(x)\leq \prod_{i=1}^{n}f_{i}(x).\tag{*}
$$
By using (*) and generalization of Hölder's inequality, I get
$$
\left(\int_{\mathbb{R}^n}|T(\overrightarrow{f})(x)w(x)|^pdx\right)^{\frac{1}{p}}\leq \left(\int_{\mathbb{R}^n}| \prod_{i=1}^{n}f_{i}(x)w(x)|^pdx\right)^{\frac{1}{p}}\leq \prod_{i=1}^{n}\left(\int_{\mathbb{R}^n}|f_{i}(x)|^{p_i}w(x)^pdx\right)^{\frac{1}{p_i}},
$$
where I apply Hölder's inequality  for the measure $d\mu(x)=w(x)^p$.
But the paper which I read now writes this inequality as following
$$
\left(\int_{\mathbb{R}^n}|T(\overrightarrow{f})(x)w(x)|^pdx\right)^{\frac{1}{p}}\leq \left(\int_{\mathbb{R}^n}| \prod_{i=1}^{n}f_{i}(x)w(x)|^pdx\right)^{\frac{1}{p}}\leq \prod_{i=1}^{n}\left(\int_{\mathbb{R}^n}|f_{i}(x)w(x)|^{p_i}dx\right)^{\frac{1}{p_i}}.
$$
Which one is true? 
 A: 
Your result is absolutely true:

Here is a way to check it: Let $\mu $ be the  measure $$d\mu(x)= w^p(x)dx$$
Now we apply the Holder inequality w.r.t the measure $\mu$ then we have, 
$$\left(\int_{\mathbb{R}^n}|T(\overrightarrow{f})(x)|^p\color{red}{d\mu(x)}\right)^{\frac{1}{p}}\leq \left(\int_{\mathbb{R}^n}| \prod_{i=1}^{n}f_{i}(x)|^p\color{red}{d\mu(x)}\right)^{\frac{1}{p}}\leq \prod_{i=1}^{n}\left(\int_{\mathbb{R}^n}|f_{i}(x)|^{p_i}\color{red}{d\mu(x)}\right)^{\frac{1}{p_i}},$$
replacing $\mu $ by its value we get: $$\left(\int_{\mathbb{R}^n}|T(\overrightarrow{f})(x)w(x)|^pdx\right)^{\frac{1}{p}}\leq \left(\int_{\mathbb{R}^n}\color{red}{\left|\prod_{i=1}^{n}f_{i}(x)\right|^p} w^p(x)dx\right)^{\frac{1}{p}}\leq \prod_{i=1}^{n}\left(\int_{\mathbb{R}^n}|f_{i}(x)|^{p_i}w(x)^pdx\right)^{\frac{1}{p_i}},$$
Which provide your result. 
Warning: However, as $\color{blue}{xa_1\times xa_2\times xa_3 = x^3\times a_1\times a_2\times a_3}$,   note that there might be confusion between the following: $$\left|\prod_{i=1}^{n}f_{i}(x)w(x)\right|^p = w^{\color{blue}{np}}(x)\left|\prod_{i=1}^{n}f_{i}(x)\right|^p$$
and $$w^p(x)\left|\prod_{i=1}^{n}f_{i}(x)\right|^p =\left|\prod_{i=1}^{n}f_{i}(x)\right|^p\times w^p(x)$$
A: Use that:
$\left(\int_{\mathbb{R}^n}| \prod_{i=1}^{n}f_{i}(x)w(x)|^pdx\right)^{\frac{1}{p}}\leq \prod_{i=1}^{n}\left(\int_{\mathbb{R}^n}|f_{i}(x)|^{p_i}w(x)^pdx\right)^{\frac{1}{p_i}}$
We can rename $\tilde{f}_i(x) = f_i(x)w(x)$ and $\tilde{w}(x) \equiv 1$ and get
$\left(\int_{\mathbb{R}^n}| \prod_{i=1}^{n}f_{i}(x)w(x)|^pdx\right)^{\frac{1}{p}} \\
= \left(\int_{\mathbb{R}^n}| \prod_{i=1}^{n}\tilde{f}_{i}(x)\tilde{w}(x)|^pdx\right)^{\frac{1}{p}} \\
\leq \prod_{i=1}^{n}\left(\int_{\mathbb{R}^n}|\tilde{f}_{i}(x)|^{p_i}\tilde{w}(x)^pdx\right)^{\frac{1}{p_i}} \\
= \prod_{i=1}^{n}\left(\int_{\mathbb{R}^n}|f_{i}(x)w(x)|^{p_i}dx\right)^{\frac{1}{p_i}} $
Edit: So I'd say both inequalities are true.
