Prove that $\sqrt{2} + \sqrt{3} + \sqrt{5}$ is irrational.

Let's argue indirectly...

Assume that $\sqrt{2} + \sqrt{3} + \sqrt{5}$ is rational.

As such, we can find a rational number $r>0$ where $$\sqrt{2} + \sqrt{3} + \sqrt{5} = r$$ Suppose, following a natural instinct, we try to reduce the number of radicals in this equation. We can start by pulling the $\sqrt{5}$ to one side and then squaring both sides...

$$\begin{align}\sqrt{2} + \sqrt{3} &= r - \sqrt{5}\\
(\sqrt{2} + \sqrt{3})^2 &= (r - \sqrt{5})^2\\
2 + 2\sqrt{6} + 3 &= r^2 - 2r\sqrt{5} +5\\
2\sqrt{6} &= r^2 - 2r\sqrt{5}\\
(2\sqrt{6})^2 &= (r^2 - 2r\sqrt{5})^2\\
24 &= r^4 - 4r^3\sqrt{5} + 20r^2\\
4r^2\sqrt{5} &= r^4 +20r^2 - 24\\
\sqrt{5} &= \frac{r^4 +20r^2 - 24}{4r^2}\end{align}$$

Recall, non-zero rational numbers are closed with respect to addition, subtraction, multiplication, and division -- so $\dfrac{r^4 +20r^2 - 24}{4r^2}$ must be a rational number. But then, $\sqrt{5}$ must be rational. We know this can't be! We have our desired contradiction. As such, we must reject our assumption.

Instead, its opposite must be true: $\sqrt{2} + \sqrt{3} + \sqrt{5}$ is irrational.