# 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 $AB$ 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.

Now raise perpendiculars $PQ$ and $OT$ to $AB$. Notice that $OT$ 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 $PQ$ 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.

Title proof of arithmetic-geometric means inequality ProofOfArithmeticgeometricMeansInequality 2013-03-22 14:49:14 2013-03-22 14:49:14 mathcam (2727) mathcam (2727) 10 mathcam (2727) Proof msc 26D15