A monic polynomial with root √2 + √3 + √5

1. Let $t=\sqrt2+\sqrt3$. Then $t^2=5+2\sqrt6$, so $(t^2-5)^2=24$, i.e.
$$t^4-10t^2+1=0.$$

2. Let $x=\sqrt2+\sqrt3+\sqrt5=t+\sqrt5$, so $t=x-\sqrt5$. Substituting:
$$(x-\sqrt5)^4-10(x-\sqrt5)^2+1=0.$$
Expand: $(x-\sqrt5)^4=x^4-4\sqrt5x^3+30x^2-20\sqrt5x+25$ and $-10(x-\sqrt5)^2=-10x^2+20\sqrt5x-50$. The terms in $x$ cancel and
$$x^4+20x^2-24=4\sqrt5\,x^3.$$
Squaring: $(x^4+20x^2-24)^2=80x^6$, that is
$$x^8-40x^6+352x^4-960x^2+576=0.$$
So $s$ is a root of this monic integer polynomial of degree $8$.

3. By the rational root theorem, a rational root of this monic polynomial is an integer. But $5.37<s<5.40$: $s$ lies strictly between $5$ and $6$ and is not an integer. Hence $s$ is irrational. $\blacksquare$

Remark. The degree $8$ is the best possible: the $8$ numbers $\pm\sqrt2\pm\sqrt3\pm\sqrt5$ are the roots of this polynomial.

Lesson. To find a polynomial killing a sum of square roots, eliminate the radicals one at a time.