AM–GM inequality (two numbers)

Start from a fact that is always true: any real square is nonnegative. Apply this to $\sqrt{a}-\sqrt{b}$, which is a real number since $a,b\ge 0$: $(\sqrt{a}-\sqrt{b})^2 \ge 0$.

Expand the square using $(u-v)^2=u^2-2uv+v^2$ with $u=\sqrt a$, $v=\sqrt b$: this gives $a - 2\sqrt{ab} + b \ge 0$, i.e. $a - 2\sqrt{ab} + b \ge 0$.

Rearrange by adding $2\sqrt{ab}$ to both sides and then dividing everything by $2$ (a positive number, so the direction is preserved): $\dfrac{a+b}{2} - \sqrt{ab} \ge 0$, which is exactly $\dfrac{a+b}{2} \ge \sqrt{ab}$.

Equality holds throughout only when the very first step was equality, i.e. $(\sqrt a-\sqrt b)^2=0$, which happens exactly when $\sqrt a=\sqrt b$, i.e. $a=b$.