A short geometrical proof can be given for the case $n=2$ of the arithmetic-geometric means inequality.

Let $a$ and $b$ be two non negative numbers. Draw the line $AB$ such that $AP$ has length $a$ , and $PB$ has length $b$ , as in the following picture, and draw a semicircle with diameter $AB$ . Let $O$ be the center of the circle.

This special case can also be proved using rearrangement inequality. Let $a,b$ non negative numbers, and assume $a\leq b$ . Let $x_1=\sqrt{a},x_2=\sqrt{b}$ , and then $x_1\leq x_2$ . Now suppose $y_1$ and $y_2$ are such that one of them is $x_1$ and the other is $x_2$ . Rearrangement inequality states that $x_1y_1+x_2y_2$ is maximum when $y_1\leq y_2$ and $x_1\leq x_2$ . So, we have$$x_1x_2+x_2x_1 \leq x_1^2 + x_2^$$ and substituting back $a,b$ gives$$2 \sqrt{ab} \leq (\sqrt{a})^2 + (\sqrt{b})^2 = a+$$ where it follows the desired result.

One more proof can be given as follows. Let $x=\sqrt{a}, y =\sqrt{b}$ . Then $(x-y)^2 \geq 0$ , and equality holds only when $x=y$ . Then, $x^2-2xy+ y^2 \geq 0$ becomes$$x^2 + y^2 \geq 2 x$$ and substituting back $a,b$ gives the desired result as in the previous proof.