# 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?