Let's say you have an inequality: $$ a^2 > b \cdot c \tag{1} $$ Let's say $a = 4$, $b = 1$, and $c = -3$ we can compute the result $$ 4^2 > 1 \cdot (-3) \tag{2} $$ so $$ 16 > -3 \tag{3} $$
Which is true so these values are solutions to the equation at (1). Now lets say we took the square root of both sides $$ a > \sqrt{b \cdot c} \tag{4} $$
Trying to put in the values for $a$, $b$, and $c$ we stated above we get $$ 4 > \sqrt{-3} \tag{5} $$
So I'm not really sure where we go from here. The square root of the negative makes me think this is a complex number like seen below $$ 4 > \sqrt{3}i \tag{6} $$
But now we have an inequality between a real number and a complex number (which we know must be true from (3)). How do we know that this is a true statement, and in general if you have $x \in \mathbb{R}$ and $y \in \mathbb{C}$ how do you determine if $x > y$ is true, $x < y$ is true, and $x = y$ is true? Thanks.