Derivation of Inequality of arithmetic and geometric means using a circle

Question

$$a,b:=\text{positive numbers}\tag{1}$$

I want to derive the following inequality.

$$\underbrace{\sqrt{ab}\leq{a+b\over 2}}_{\text{Inequality of arithmetic and geometric mean}}\tag{2}$$

To derive it, I've reffered the following from here.

Of course I can get$~{a+b\over 2}~$of$~\text{AO}~$since it is a radius of diameter$~a+b~$

But how can I get$~\sqrt{ab}?~$

I tried Pythagorean theorem to apply it here but didn't work.

$$ \begin{cases} c:=\text{PG}\\ d:=\text{GR}\\ l:=\text{GQ} \end{cases}\tag{3} $$

$$a^2+l^2=c^2\tag{4}$$

$$b^2+l^2=d^2\tag{5}$$

$$\therefore~~c^2-a^2=d^2-b^2\tag{6}$$

Answer 1

• $\triangle PQG$ and $\triangle PGR$ are similar since they both have $\angle QPG$ and a right angle
• $\triangle GQR$ and $\triangle PGR$ are similar since they both have $\angle QRG$ and a right angle
• so $\triangle PQG$ and $\triangle GQR$ are similar
• so $\frac{|QP|}{|QG|}=\frac{|QG|}{|QR|}$
• i.e. $|QG|^2=|QP|\,|QR|=ab$
• and thus $|QG|=\sqrt{ab}$