Let a = $\sqrt{5}+\sqrt{7}$ and b = $\sqrt{12}+\sqrt{3}$. Which is greater?

The correct option is a < b.

We are given two expressions:
$a=\sqrt{5}+\sqrt{7}$
$b=\sqrt{12}+\sqrt{3}$

To compare $a$ and $b$, we can square both positive numbers and compare their squares.

First, let us find $a^2$ using the identity ${(x+y)}^2=x^2+y^2+2xy$:

$a^2={(\sqrt{5}+\sqrt{7})}^2$
$a^2={\left(\sqrt{5}\right)}^2+{\left(\sqrt{7}\right)}^2+2·\sqrt{5}·\sqrt{7}$
$a^2=5+7+2\sqrt{35}$
$a^2=12+2\sqrt{35}$

Next, let us find $b^2$ using the same algebraic identity:

$b^2={(\sqrt{12}+\sqrt{3})}^2$
$b^2={\left(\sqrt{12}\right)}^2+{\left(\sqrt{3}\right)}^2+2·\sqrt{12}·\sqrt{3}$
$b^2=12+3+2\sqrt{36}$
$b^2=15+2·6$
$b^2=15+12=27$

Alternatively, we can express $b^2$ as:
$b^2=12+3+2\sqrt{36}=12+2\sqrt{36}+3$

Now, comparing the square expressions:
$a^2=12+2\sqrt{35}$
$b^2=12+2\sqrt{36}+3$

Since $\sqrt{35}<\sqrt{36}$, it follows that:
$2\sqrt{35}<2\sqrt{36}$

Therefore:
$12+2\sqrt{35}<12+2\sqrt{36}+3$

This gives:
$a^2<b^2$

Since both $a$ and $b$ are positive real numbers, taking the square root on both sides yields:
$a<b$