A simple inequality and root I need to use the following inequality
$$ \left(\sum_{i=1}^N X_i \right)^{1/2} \leq \left( \sum_{i=1}^N X_i^{1/2} \right),\hspace{0.5cm} 0 \leq X_i $$
But i can't remember its name.
Is the inequality correct?
If it's correct, then how can i prove it?
 A: I don't now about a name but this is a consequence of the triangle inequality. 
The triangle inequality states that for any vectors $z_1,y \in \mathbb{R}^n$, we have:
$$\|z_1+y\| \leq \|z_1\|+\|y\|$$
Let $y=z_2+z_3$ then,
$$\|z_1+z_2+z_3 \| \leq \|z_1\|+\|z_2+z_3\| \leq \|z_1\|+\|z_2\|+\|z_3\|$$
Continuing in this fashion we have,
$$\| \sum_{k=1}^n z_k\| \leq \sum_{k=1}^n \|z_k\|$$
This basically proves it. To finish consider vectors $z_1=\langle \sqrt{x_1},0,\ldots,0 \rangle \in \mathbb{R}^n$, and $z_2=\langle 0, \sqrt{x_2},0,\ldots,0 \rangle \in \mathbb{R}^n$, etc. Here there are zeroes everywhere except in the spot indicated by the index $k$. In the spot with the $k$-th index lies $\sqrt{x_k}$.

Then we have,
$$\| \sum_{k=1}^n z_k \|=\| \langle \sqrt{x_1},\sqrt{x_2},\ldots,\sqrt{x_n} \rangle \|$$
$$=\sqrt{x_1+x_2+\cdots+x_n}$$
Whereas we have,
$$\sum_{k=1}^{n} \| z_k\|=\sqrt{x_1}+\sqrt{x_2}+\cdots+\sqrt{x_n}$$
So we conclude,
$$\sqrt{x_1+x_2+\cdots+x_n}  \leq \sqrt{x_1}+\sqrt{x_2}+\cdots+\sqrt{x_n}$$
A: $$\sum_{i=1}^nX_i^{\frac{1}{2}}=\sqrt{\left(\sum_{i=1}^nX_i^{\frac{1}{2}}\right)^2}=\sqrt{\sum_{i=1}^nX_i+2\sum_{1\leq i<j\leq n}^nX_i^{\frac{1}{2}}X_j^{\frac{1}{2}}}\geq\sqrt{\sum_{i=1}^nX_i}$$
