HUNT THE CHALLANGES OF MATHS!!!! ~2

Squaring both sides gives a + b + 2 sqrt(ab) = c. But, using triangle inequality for a + b > c = a + b + 2 sqrt(ab) implies that ab < 0, which is not a length of a triangle...

WLOG, let us take $0 < a \le b \le c$ . Taking the Law of Cosines (with $\theta$ the angle opposite of side length $c$ ), we have:

$c^2 = a^2 + b^2 - 2ab \cdot \cos\theta$ ;

or $(\sqrt{a}+\sqrt{b})^4 = a^2 + b^2 - 2ab \cdot \cos\theta$ ;

or $a^2 + 4a^{3/2}b^{1/2} + 6ab + 4a^{1/2}b^{3/2} + b^2 = a^2 + b^2 - 2ab \cdot \cos\theta$ ;

or $4\sqrt{ab}(a+b) + 6ab = -2ab \cdot \cos\theta;$

or $\boxed{-\frac{2(a+b)}{\sqrt{ab}} - 3 = \cos\theta}.$

Since $-1 < \cos\theta < 1$ must be satisfied for a triangle to exist, we now calculate the following inequalities:

$-\frac{2(a+b)}{\sqrt{ab}} - 3 < 1 \Rightarrow -2(a+b) < 4\sqrt{ab} \Rightarrow a^2 + 2ab + b^2 < 4ab \Rightarrow a^2-2ab+b^2 < 0 \Rightarrow (a-b)^2 < 0$ (CONTRADICTION for $a,b \in \mathbb{R^{+}}$ ).

$-\frac{2(a+b)}{\sqrt{ab}} - 3 > -1 \Rightarrow 2\sqrt{ab} < -2(a+b) \Rightarrow \sqrt{ab} < -(a+b)$ (CONTRADICTION for $a,b \in \mathbb{R^{+}}$ ).

Hence, no triangles can be formed under the given side length condition $\sqrt{a} + \sqrt{b} = \sqrt{c}.$

It is a general rule that the sum of the lengths of 2 sides of a triangle must be greater than that of the other. In other words, a + b > c, where a, b, and c are lengths of sides of a triangle