A Proof for the Inequality of Arithmetic and Geometric Means

**bkarpuz** · November 9th 2010, 07:48 AM

Dear friends,

while working on some computations, I figured out that my arguments supply a proof for
the arithmetic and geometric mean inequality by using calculus. I wanted to share here.

Assertion. For all $x,y\geq0$ show that $\dfrac{x+y}{2}\geq\sqrt{xy}$ with equality if and only if $x=y$ .
Proof. The case $xy=0$ is trivial.
Let $x,y>0$ and define
$f(x,y):=\dfrac{x+y}{2}-\sqrt{xy}$ .
We shall prove that for all $x,y>0$ , $f(x,y)\geq0$ with equality if and only if $x=y$ .
In this case, we may find $\lambda>0$ such that $y=\lambda x$ (in particular, $\lambda:=y/x$ ), which yields
$f(x,\lambda x)=g(\lambda)x$ , where $g(\lambda):=\dfrac{\lambda+1}{2}-\sqrt{\lambda}$ for $\lambda>0$ .
As $x>0$ , we have to show that for all $\lambda>0$ , $g(\lambda)\geq0$ with equality if and only if $\lambda=1$ .
We can compute that
$g^{\prime}(\lambda)=\dfrac{1}{2}\bigg(1-\dfrac{1}{\sqrt{\lambda}}\bigg)$ and $g^{\prime\prime}(\lambda)=\dfrac{1}{4\sqrt{\lambda ^{3}}}>0$ for all $\lambda>0$ .
The function $g$ is therefore convex and this shows that $g(1)=0$ is the global minimum value
since $g^{\prime}(1)=0$ . Finally, for all $\lambda>0$ , $g(\lambda)\geq g(1)=0$ with equality if and only if $\lambda=1$ . $\rule{0.2cm}{0.2cm}$

**chisigma** · November 9th 2010, 08:50 AM

An alternative is the use of polar coordinates...

$x= \rho\ \cos \theta$

$y= \rho\ \sin \theta$ (1)

... so that is...

$\displaystyle \frac{x+y}{2} = \frac{\rho}{2} (\sin \theta + \cos \theta)$

$\displaystyle \sqrt{x\ y} = \rho\ \sqrt{\sin \theta \ \cos \theta}$ (2)

Now the proof of the 'inequality' is equivalent to prove that is...

$\displaystyle \frac{1}{4} \ge \frac{\sin \theta \ \cos \theta}{2}$ (3)

... and (3) is true because the function $f(\theta) = \sin \theta \ \cos \theta$ has its maximum for $\theta= \frac{\pi}{4}$ and is $f(\frac{\pi}{4}) = \frac{1}{2}$ ...

Kind regards

$\chi$ $\sigma$

**Plato** · November 9th 2010, 09:29 AM

Here is an easy proof.
It should be clear that $(a-b)^2\ge 0$ .
So that $a^2+b^2\ge 2ab$ .
If $x\ge 0~\&~y\ge 0$ then let $a=\sqrt{x}~\&~ a=\sqrt{y}$ .

**Krizalid** · November 9th 2010, 01:28 PM

Originally Posted by chisigma

$\displaystyle \frac{1}{4} \ge \frac{\sin \theta \ \cos \theta}{2}$ (3)

$\displaystyle\frac{\sin (\theta )\cos \theta }{2}\le \frac{1}{4}\implies \sin (2\theta )\le 1.$

**bkarpuz** · November 9th 2010, 01:45 PM

@bkarpuz: I should take $\lambda^{2}$ instead of $\lambda$ .

@chisigma: This makes use of $\lambda=\tan(\theta)$ .
@Plato: I know that that is the most famous and easiest proof. ^^
@Krizalid: Nice trick!

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