How to prove an inequality by using a mathematical induction?

I have to prove the following:

$$ sqrt {x_1} + sqrt {x_2} + … + sqrt {x_n} ge sqrt {x_1 + x_2 + … + x_n} $$

For each $ n ge $ 2 and $ x_1, x_2, …, x_n in Bbb R $

Here is my attempt:

Consider $ P (n): sqrt {x_1} + sqrt {x_2} + … + sqrt {x_n} ge sqrt {x_1 + x_2 + … + x_n} $

$$ P (2): sqrt {x_1} + sqrt {x_2} ge sqrt {x_1 + x_2} $$
$$ x_1 + x_2 + 2 sqrt {x_1x_2} ge x_1 + x_2 $$
Which is true because $ 2 sqrt {x_1x_2}> $ 0.

$$ P (n + 1): sqrt {x_1} + sqrt {x_2} + … + sqrt {x_n} + sqrt {x_ {n + 1}} ge sqrt {x_1 + x_2 + … + x_n + x_ {n + 1}} $$

From the hypothesis we have:

$$ sqrt {x_1} + sqrt {x_2} + … + sqrt {x_n} + sqrt {x_ {n + 1}} ge sqrt {x_1 + x_2 + … + x_n} + sqrt {x_ {n + 1}} ge sqrt {x_1 + x_2 + … + x_n + x_ {n + 1}} $$

Squaring on both sides of the right side:

$$ x_1 + x_2 + … + x_n + x_ {n + 1} + 2 sqrt {x_ {n + 1} (x_1 + x_2 + … + x_n)} ge x_1 + x_2 + … + x_n + x_ {n + 1} $$

Which is true, so $ P (n + 1) $ is true too.

I do not know if I did it correctly?