Find the minimum value of
\[\frac{x^2 + 7}{\sqrt{x^2 + 3}}\]over all real numbers $x.$
I think we can simplify this expression by letting $y = x^2 + 3$. So then the expression we're trying to minimize becomes $\frac{y + 4}{\sqrt{y}}$. And $\frac{y + 4}{\sqrt{y}} = \sqrt{y} + \frac{4}{\sqrt{y}}$. So then we need to minimize $\sqrt{y} + \frac{4}{\sqrt{y}}$. The derivative of $\sqrt{y}$ is $\frac{1}{2 \sqrt{y}}$, and the derivative of $\frac{4}{\sqrt{y}}$ is $-\frac{4}{2y\sqrt{y}} = -\frac{2}{y\sqrt{y}}$. So then the derivative of $\sqrt{y} + \frac{4}{\sqrt{y}}$ is $\frac{1}{2 \sqrt{y}} - \frac{2}{y\sqrt{y}}$ Setting that equal to zero, we get $\frac{1}{2 \sqrt{y}} - \frac{2}{y\sqrt{y}} = 0$ Which is equivalent to $\frac{1}{2 \sqrt{y}} = \frac{2}{y\sqrt{y}}$. Cross-multiplying, we find that $y = 4$. So then $x^2 + 3 = 4$, which means that $x^2 = 1$. So the only possible values of $x$ are $1$ and $-1$. The original expression is $\frac{x^2 + 7}{\sqrt{x^2 + 3}}$, so if $x = 1$, the value is $\frac{1^2 + 7}{\sqrt{1^2 + 3}} = \frac{8}{2} = 4$. If $x = -1$, the value is $\frac{(-1)^2 + 7}{\sqrt{(-1)^2 + 3}} = \frac{8}{2} = 4$. So the minimum value is $4$.

