How to show that the integral inequality holds for vector-valued functions. [duplicate]

If I define the integral for $\int_a^b\textbf{F}(t)dt$ as $(\int_a^b F_1(t)dt, \int_a^bF_2(t)dt,\ldots, \int_a^bF_n(t)dt )$. How do I then show that

$$\left|\int_a^b\textbf{F}(t)dt \right|\le\int_a^b\left|\textbf{F}(t) \right|dt?$$

If we write out the inequality we have

$$\sqrt{\left(\int_a^b F_1(t)dt\right)^2+\left(\int_a^bF_2(t)dt\right)^2+\cdots+ \left(\int_a^bF_n(t)dt\right)^2 }\le\int_a^b\sqrt{F_1(t)^2+F_2(t)^2+\cdots+F_n(t)^2}dt.$$

attempt:

If I use Jenssen's inequality I get for the left side: $$\sqrt{\left(\int_a^b F_1(t)dt\right)^2+\left(\int_a^bF_2(t)dt\right)^2+\cdots+ \left(\int_a^bF_n(t)dt\right)^2 }\le\sqrt{\int_a^b F_1^2(t)dt+\int_a^bF_2^2(t)dt+\cdots+ \int_a^bF_n^2(t)dt }\\=\sqrt{\sum\limits_{i=1}^n\int_a^bF_i^2(t)dt}=\sqrt{\int_a^b\sum\limits_{i=1}^nF_i^2(t)dt}.$$

The problem I have now is that the square root function is concave, so if I use Jenssen again, I get the opposite inequality as the one I need. Any idea on how to solve this?

The linearity of the integral, that of the inner product and the Schwarz inequality imply, for any $\vec v\in \mathbb R^n,$

$\tag 1\left \langle \int_a^bF(t)dt,\vec v\right \rangle=\int_a^b\langle F(t),\vec v\rangle dt\le \int_a^b|\langle F(t),\vec v\rangle| dt\le \int_a^b|F(t)|\vec v| |dt\le |\vec v|\int_a^b |F(t)|dt .$

$\tag 2\text{Taking}\quad \vec v=\int_a^bF(t)dt,\quad \text{we get}\quad |\vec v|^2\le |\vec v|\int_a^b |F(t)|dt$

$\tag3 \text{which is}\quad \left |\int_a^bF(t)dt\right |\le \int_a^b |F(t)|dt.$