Let $a=\sqrt{5}+\sqrt{7}$ and $b=\sqrt{12}+\sqrt{3}$. Which is greater?
$a=b$
$a>b$
$a<b$
Cannot be determined
Correct answer: $a<b$
Solution
We have $a=\sqrt{5}+\sqrt{7}$ and $b=\sqrt{12}+\sqrt{3}=2\sqrt{3}+\sqrt{3}=3\sqrt{3}$. Numerically, $a\approx 2.236+2.646=4.882$ and $b\approx 5.196$, so $a<b$.