Between which two consecutive integer numbers is $\sqrt{2}+\sqrt{3}$? Between which two consecutive integer numbers is $\sqrt{2}+\sqrt{3}$?
My thoughts: $\sqrt{2}$ is $\approx1,4$ and $\sqrt{3}$ is $\approx{1,7}$ so their sum must be of the interval $(3;4)$. Any more strict approaches? Thank you in advance!
 A: First we have $\sqrt{2} + \sqrt{3} < \sqrt{4} + \sqrt{4} = 4$. To show that $\sqrt{2} + \sqrt{3} > 3$, we square both sides, and see that it suffices to show that $5 + 2\sqrt{6} > 9$. But this follows from $\sqrt{6} > 2$.
A: You are entirely right. To be more precise, you could say that
$$
1.4<\sqrt2<1.5\\
1.7<\sqrt3<1.8
$$
and therefore
$$
3.1<\sqrt2+\sqrt3<3.3
$$
Alternately, to keep all approximations integral (and this is more likely the intended solution), we could also study $(\sqrt2+\sqrt3)^2=2+2\sqrt6+3$. Noting that $2<\sqrt6<3$ gives
$$
9<2+2\sqrt6+3<11<16
$$
which again means $3<\sqrt2+\sqrt3<4$.
A: I think they expect you to do something like this:
Since $2,3<4$ we know $\sqrt2,\sqrt3<2$, so $\sqrt2+\sqrt3<4$.
On the other hand, $(\sqrt2+\sqrt3)^2=2+3+2\sqrt6>9$ since $\sqrt6>\sqrt4=2$.  This shows that $\sqrt2+\sqrt3>3$ so we have $$
3<\sqrt2+\sqrt3<4$$
A: This first step (bracketing between plausible consecutive integers) was necessary.
Now we have to prove in a rigorous way that
$$\sqrt{2}+\sqrt{3}>3 \tag{1}$$ and
$$\sqrt{2}+\sqrt{3}<4 \tag{2}$$
Let us establish (2).
As function $x\to x^2$ is strictly increasing on $(0,\infty)$, therefore bijective, (2) is equivalent to:
$$(\sqrt{2}+\sqrt{3})^2<4^2  \ \iff  \ 5+2 \sqrt{6}<16  \ \iff  \ \sqrt{6}<\frac{11}{2} $$
itself equivalent, for the same reason, to:
$$6<\frac{121}{4} \ \iff 24 \ < \ 121$$
which is true.
Proceed in the same way for inequation (1).
