Does $\sqrt{a+b}\leq \sqrt{a}+\sqrt{b}$?

Question

In the case of $$\sqrt{(x_n-\ell_1)+(y_n-\ell_2)}\leq \sqrt{(x_n-\ell_1)^2} + \sqrt{(y_n-\ell_2)^2} = |x_n-\ell_1|+|y_n-\ell_2|$$

it is true, if we take the rise the two sides in the power of $2$ we get:

\begin{align} & (x_n-\ell_1)+(y_n-\ell_2)\leq \left( \sqrt{(x_n-\ell_1)^2}+\sqrt{(y_n-\ell_2)^2} \,\right)^2 \\[10pt] = {} &{(x_n-\ell_1)+(y_n-\ell_2)}+\sqrt{(x_n-\ell_1)+(y_n-\ell_2)} \end{align}

Does the same is true for $$\sqrt{a+b}\leq \sqrt{a}+\sqrt{b} \text{ ?}$$

doesn't $\sqrt{a+b}$ is a product of $3$ components? $\sqrt{a}+\sqrt{b}+\text{(something positive)}$ and therefore $$\sqrt{a+b}\geq \sqrt{a}+\sqrt{b} \text{ ?}$$

Answer 1

Quite simply, the answer to your question from the title is "yes", if we assume that $a$ and $b$ are non-negative real numbers.

Since both of the sides are positive, we can square the entire equation, and we get a trivial inequality: $$ \sqrt{a+b} \leq \sqrt{a} + \sqrt{b} \iff a+b \leq a+b+2\sqrt{ab} \iff 0 \leq 2\sqrt{ab}. $$

Answer 2

Hint: Square both sides.${}{}$