## 48. Math Puzzle $5 / 69$

Which sum is greater: $\sqrt{7}+\sqrt{10}$ or $\sqrt{3}+\sqrt{19}$?
Both expressions are squared:

$$
(\sqrt{7}+\sqrt{10})^{2}=17+2 \sqrt{70} \quad \text { and } \quad(\sqrt{3}+\sqrt{19})^{2}=22+2 \sqrt{57}
$$

We subtract 17 from both values. The remaining values are squared again:

$$
(2 \sqrt{70})^{2}=280 \quad \text { and } \quad(5+2 \sqrt{57})^{2}=253+20 \sqrt{57}
$$

We subtract 253 from each, resulting in 27 and $20 \sqrt{57}$. Since $\sqrt{57}$ is greater than 2, $20 \sqrt{57}>40$ and consequently $\sqrt{7}+\sqrt{10}<\sqrt{3}+\sqrt{19}$.