Geometric and Arithmetic Mean

I don’t know why I always have trouble proving this.

$x,y \geq 0$ , $\sqrt{xy} \leq \frac{x+y}{2}$ , with equality if and only if $x = y$

There are a lot of proofs for this that I consider to be ugly because of the type of algebraic gimmickry that they use. The best proof that I could think of is as follows.

$0 \leq (x - y)^2$ (with equality if and only if $x = y$ )

$0 \leq x^2 - 2xy + y^2$

$0 \leq x^2 + 2xy - 4xy + y^2$ (1.1)

$4xy \leq x^2 + 2xy + y^2$

$4xy \leq \left(x+y\right)^{2}$

Taking positive square roots, and by the strict monotonicity of the positive square root function,

$2\sqrt{xy} \leq (x + y)$

$\sqrt{xy} \leq \frac{(x + y)}{2}$

Note that, the strict monotonicity of the positive square root function is required for the equality condition.

On the other hand, if in (1.1) if we said adding $4xy$ to both sides, this would have become an ugly proof. I “have a low tolerance” for algebraic gimmickry like that.

Here’s another ugly proof:

We have $x^2 - 2xy + y^2 = \left( x - y \right)^{2} \geq 0$ (1.2) with equality if and only if $x = y$ . This yields

$xy \leq xy + \frac{1}{4} \left( x^2 - 2 xy + y^2 \right)$ (1.3)

$= xy+\frac{1}{4}x^{2}-\frac{1}{2}xy+\frac{1}{4}y^{2}$

$= \frac{1}{4}x^{2}+\frac{1}{2}xy+\frac{1}{4}y^{2}$

$= \frac{1}{4}\left(x^{2}+2xy+y^{2}\right)$

$= \frac{1}{4}\left(x+y\right)^{2}$

Taking roots yields the desired inequality. It is clar that we have equality if and only if the second summand in (1.3) vanishes; by (1.2) this is only possible only if $x = y$ .

(This proof was taken form Fall 2004 and Winter 2005 notes by Volker Runde)

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Ramblings of an Insane Idiot

Geometric and Arithmetic Mean

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