real analysis – Prove that $ left | int_X mathbf {f} d mu right | _p leq int_X left | mathbf {f} right | _p d mu $ for $ mathbf {f} = (f_1, …, f_n) $

I have problems with the following proof.

Define $ mathbf {f} (x) = (f_1 (x), …, f_n (x)) $ Where $ f_i: X to mathbb {R} $ for every positive integer $ i leq n $, and each one $ f_i $ is integrable.

Prove that
$$ left | int_X mathbf {f} \ mathrm {d} mu right | _p leq int_X left | mathbf {f} right | _p \ mathrm {d} mu $$

to the $ 1 leq p leq infty $

Now we can explicitly write the left side as

$$ left | int_X mathbf {f} \ mathrm {d} mu right | _p = left ( sum_ {j = 1} ^ n left | int_X f_j \ mathrm {d} mu right | ^ p right) ^ {1 / p} $$

Now the exponent annoys me. If we had the case $ p = 1 $, then it's just there for every integrable function $ g: X to mathbb {R} $, we have

$$ left | int_X g \ mathrm {d} mu right | leq int_X | g | \ mathrm {d} mu $$

Consequently,

$$ left | int_X mathbf {f} \ mathrm {d} mu right | = sum_ {j = 1} ^ n left | int_X f_j mathrm {d} mu right | leq sum_ {j = 1} ^ n int_X | f_j | \ mathrm {d} mu = int_X left ( sum_ {j = 1} ^ n | f_j | right) \ mathrm {d} mu = int_X | mathbf {f} | \ mathrm {d} mu $$

But I can not handle the exponential $ 1 <p < infty $

Thanks!