Here is an alternative proof, which is circuitous but perhaps enjoyable as a curiosity:
Fact: For any fixed $P > 0$, the rectangle of perimeter $P$ with maximal area is a square.
Consequence: Given a $2a\times 2b$ rectangle, it has area $4ab$ and perimeter $4a + 4b$. But the square of the same perimeter has side-length $a+b$, hence area $(a+b)^2$.
The Fact above now provides the desired inequality in terms of square and rectangle areas:
$$(a+b)^2 \geq 4ab $$
Fact Proof (for completeness): Let us denote one side of the rectangle by $x$ and the other by $y$. Then the perimeter is $P = 2x + 2y$, so that $2y = P - 2x$, whence $y = P/2 - x$.
Now the area is $xy = x(P/2 - x) = Px/2 - x^2$. To maximize the last expression, observe that it is a downward facing parabola; so its maximum occurs at the vertex, i.e., $P/4$.
Thus, $x = P/4 = y$ as claimed.