proof of arithmetic-geometric means inequality

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 $A B$ such that $A P$ has length $a$ , and $P B$ has length $b$ , as in the following picture, and draw a semicircle with diameter $A B$ . Let $O$ be the center of the circle.

Now raise perpendiculars $P Q$ and $O T$ to $A B$ . Notice that $O T$ is a radius, and so

OT=\\frac{AB}{2}=\\frac{a+b}{2}

Also notice that $PQ\\leq OT$ for any point $P$ , and equality is obtained only when $P=O$ , that is, when $a=b$ .

Notice also that $P Q$ is a height over the hypotenuse on right triangle $\\triangle AQB$ . We have then triangle similarities $\\triangle AQB\\sim\\triangle APQ\\sim\\triangle QPB$ , and thus

\\frac{AP}{PQ}=\\frac{PQ}{PB}

which implies $PQ=\\sqrt{AP\\cdot PB}=\\sqrt{ab}$ . Since $PQ\\leq OT$ , we conclude

\\sqrt{ab}\\leq\\frac{a+b}{2}.

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 $x_{1}y_{1}+x_{2}y_{2}$ is maximum when $y_{1}\\leq y_{2}$ and $x_{1}\\leq x_{2}$ .So, we have

x_{1}x_{2}+x_{2}x_{1}\\leq x_{1}^{2}+x_{2}^{2}

and substituting back $a, b$ gives

2\\sqrt{ab}\\leq(\\sqrt{a})^{2}+(\\sqrt{b})^{2}=a+b

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 2xy

and substituting back $a, b$ gives the desired result as in the previous proof.