# Thread: A Proof for the Inequality of Arithmetic and Geometric Means

1. ## A Proof for the Inequality of Arithmetic and Geometric Means

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

2. An alternative is the use of polar coordinates...

$\displaystyle x= \rho\ \cos \theta$

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

... so that is...

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

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

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

4. Originally Posted by chisigma
$\displaystyle \displaystyle \frac{1}{4} \ge \frac{\sin \theta \ \cos \theta}{2}$ (3)
$\displaystyle \displaystyle\frac{\sin (\theta )\cos \theta }{2}\le \frac{1}{4}\implies \sin (2\theta )\le 1.$

5. @bkarpuz: I should take $\displaystyle \lambda^{2}$ instead of $\displaystyle \lambda$.
@chisigma: This makes use of $\displaystyle \lambda=\tan(\theta)$.
@Plato: I know that that is the most famous and easiest proof. ^^
@Krizalid: Nice trick!