DEMONSTRATION HELP

[hoosie]

 

[JeffM]

Hoosie

I'll show you a quick way to find many proofs of inequality. (I don't guarantee it will work every time.) It is an application of Proclus's method.

[MATH]0 < x < y \text { is given.}[/MATH]
[MATH]y \ ? \ \sqrt{\dfrac{x^2 + y^2}{2}} \implies y^2 \ ? \ \dfrac{x^2 + y^2}{2} \implies[/MATH]
[MATH]2y^2 \ ? \ x^2 + y^2 \implies y^2 \ ? \ x^2 \implies y \ ? \ x.[/MATH]
But y > x. So we have "solved" for ?, which stands for a relationship rather than a quantity. (You have to be careful if you multiply by a non-positive quantity or take reciprocals.) Now we see if it works in reverse. If so, we have a simple proof.

[MATH]0 < x < y \text { is given.}[/MATH]
[MATH]\therefore 0 < x^2 < y^2 \implies 0 < x^2 + y^2 < y^2 + y^2 \implies 0 < x^2 + y^2 < 2y^2 \implies[/MATH]
[MATH]0 < \dfrac{x^2 + y^2}{2} < y^2 \implies 0 < \sqrt{\dfrac{x^2 + y^2}{2}} < y.[/MATH]
Helpful, no?

 

[pka]

If \(\displaystyle 0<x<y\) then \(\displaystyle \sqrt{xy}<\dfrac{x+y}{2}\)

From the given we know:
\(\displaystyle (y-x)^2>0\\y^2-2xy+x^2>0\\x^2+2xy+y^2>4xy\\(x+y)^2>4xy\\(x+y)>2\sqrt{xy}\\\dfrac{x+y}{2}>\sqrt{xy}\)

 

[hoosie]

Like the way you did that Jeff!

Hoosie

I'll show you a quick way to find many proofs of inequality. (I don't guarantee it will work every time.) It is an application of Proclus's method.

[MATH]0 < x < y \text { is given.}[/MATH]
[MATH]y \ ? \ \sqrt{\dfrac{x^2 + y^2}{2}} \implies y^2 \ ? \ \dfrac{x^2 + y^2}{2} \implies[/MATH]
[MATH]2y^2 \ ? \ x^2 + y^2 \implies y^2 \ ? \ x^2 \implies y \ ? \ x.[/MATH]
But y > x. So we have "solved" for ?, which stands for a relationship rather than a quantity. (You have to be careful if you multiply by a non-positive quantity or take reciprocals.) Now we see if it works in reverse. If so, we have a simple proof.

[MATH]0 < x < y \text { is given.}[/MATH]
[MATH]\therefore 0 < x^2 < y^2 \implies 0 < x^2 + y^2 < y^2 + y^2 \implies 0 < x^2 + y^2 < 2y^2 \implies[/MATH]
[MATH]0 < \dfrac{x^2 + y^2}{2} < y^2 \implies 0 < \sqrt{\dfrac{x^2 + y^2}{2}} < y.[/MATH]
Helpful, no?

Yes Jeff - I like your method. Thanks