Can we proof that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$?; given $0 < a < b < 1$

Question

I have found in some test problem:

Given $0 < a < b < 1$, can we conclude that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$?

I divide the combination of a and b into a few cases. Then, see what happens.

case 1 : $a \rightarrow 0$ and $b \rightarrow 0$

When $a$ approaches to zero and $b$ approaches to $a$ (which is zero), the expression will be evaluated as $\sqrt{0 + 0}$ and $\sqrt{0} + \sqrt{0}$. Since $0 = 0$, this case cannot conclude that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$

case 2 : $a \rightarrow 1$ and $b \rightarrow 1$

When $b$ approaches to one and $a$ approches to $b$ (which is one), the expression will be evaluated as $\sqrt{1 + 1}$ and $\sqrt{1} + \sqrt{1}$. Since $\sqrt{2} < 2$, then this case concludes that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$

case 3 : $a \rightarrow 0$ and $b \rightarrow 1$

When $a$ approaches to zero and $b$ approches to one, the expression will be evaluated as $\sqrt{0 + 1}$ and $\sqrt{0} + \sqrt{1}$. Since $1 = 1$, then this case cannot concludes that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$

case 4 : $a \rightarrow b$

When $a$ approaches to $b$ the expression will be evaluated as $\sqrt{b + b}$ and $\sqrt{b} + \sqrt{b}$. Since $\sqrt{2b} < 2\sqrt{b} $, then this case concludes that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$

---

From above cases, am I still missing some point? If not, How should I write the conclusion from that pieces of thinking mathematically? Because I instinctively believe that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$ should be true (and not $\sqrt{a + b} \leq \sqrt{a} + \sqrt{b}$).

Answer 1

We know that $$0 < 2\sqrt{ab}$$

add $a+b$ to both sides, we have

$$a+b < a+2\sqrt{ab}+b=(\sqrt{a}+\sqrt{b})^2$$

then we can take square root on both sides.

Answer 2

Since $f(x)=x^2$ is strictly increasing for $x\ge 0$ we have

$$A<B \iff A^2<B^2 \quad A,B\ge 0$$

and therefore

$$\sqrt{a + b} < \sqrt{a} + \sqrt{b}\iff (\sqrt{a + b})^2 < (\sqrt{a} + \sqrt{b})^2 \iff a+b <a+b+2\sqrt{ab}$$

that is

$$2\sqrt{ab}>0$$

which is true.

Answer 3

This proof works for all $a,b\geq 0$. Consider two vectors in the $2$-dimensional Euclidean plane $\mathbb{R}^2$ (with usual Euclidean norm $\Vert\bullet\Vert$): $u=(\sqrt{a},0)$ and $v=(0,\sqrt{b})$. We have $\Vert u\Vert+\Vert v\Vert\geq\Vert u+v\Vert$ by the triangle inequality. This gives $$\sqrt{a}+\sqrt{b}\geq \sqrt{a+b}.$$ Since the equality occurs if, and only if, $u$ and $v$ are parallel, we conclude that the only equality cases are (1) $a=0$ and (2) $b=0$. Hence, if $a,b>0$, $$\sqrt{a}+\sqrt{b}>\sqrt{a+b}.$$

Answer 4

A bit of geometry.

Set $x^2 = a$; $y^2 =b$, where $x,y >0.$

Choose a.point $C(x,y)$ in the first quadrant.

Points : $O(0,0)$, $A(x,0)$ and $C(x,y)$.

$\triangle OAC$ is a right triangle with side lengths $x,y$ and $\sqrt{x^2+y^2.}$ (Pythagoras)

The sum of the lengths of two sides in a triangle is greater than the length of the third side:

$\sqrt{x^2+y^2} < x+y$, or in original

notation $\sqrt{a+b} < √a+√b.$