Proving Inequality: x1, x2, y1, y2 Non-zero Constants

[skeeterrr]

Homework Statement

Let x1, x2, y1, y2 be arbitrary non-zero constants. Let

x = x1 / root((x1)^2+(x2)^2)

y = y1 / root((y1)^2+(y2)^2)

Show that

(2(x1)(y1))/(root(x1)^2+(x2)^2)root(y1)^2+(y2)^2)) =< (x1)^2/(x1)^2+(x2)^2 + (y1)^2/(y1)^2+(y2)^2

Homework Equations

The Attempt at a Solution

Well, I get 2xy =< x^2 + y^2 by replacing that whole complex equation above.

Then it becomes similar to that other post I made...

making a contradictory statement:

2xy >= x^2 + y^2

0 >= x^2 - 2xy + y^2

0 >= (x-y)^2

but (x-y)^2 must be either 0 or a positive integer

0 =< (x-y)^2

0 =< x^2 - 2xy + y^2

2xy =< x^2 + y^2

and if I replace x and y, I get this again:

(2(x1)(y1))/(root(x1)^2+(x2)^2)root(y1)^2+(y2)^2)) =< (x1)^2/(x1)^2+(x2)^2 + (y1)^2/(y1)^2+(y2)^2

Am I doing this right? It doesn't really seem like it...

 

[praharmitra]

This question is just a very exaggerated way of saying a very simple thing

Arithmetic Mean of two numbers is greater than the Geometric Mean, i.e

(a+b)/2 >= sqrt(ab) always, for a,b > 0

just substitute(as you already have) a = x^2 and b = y^2 and u get the same inequality back

your method is correct

 

[Architeuthis Dux]

Your approach seems to be on the right track. However, instead of using a contradictory statement, you can try to prove the inequality directly by using the properties of square roots and simplifying the expression. For example, you can start by simplifying the left side of the inequality:

(2(x1)(y1))/(root(x1)^2+(x2)^2)root(y1)^2+(y2)^2))

= (2x1y1)/(root((x1)^2+(x2)^2)root((y1)^2+(y2)^2)) (using the definition of x and y)

= 2x1y1 / root((x1)^2+(x2)^2) * 1/root((y1)^2+(y2)^2) (using the property of square roots)

= 2x1y1 / root((x1)^2+(x2)^2) * root((y1)^2+(y2)^2) / ((y1)^2+(y2)^2) (multiplying by 1 in the form of root((y1)^2+(y2)^2) / root((y1)^2+(y2)^2))

= 2x1y1 / root((x1)^2+(x2)^2) * root((x1)^2+(x2)^2 + (y1)^2+(y2)^2 - (x1)^2-(x2)^2) / ((y1)^2+(y2)^2) (adding and subtracting (x1)^2+(x2)^2)

= 2x1y1 / root((x1)^2+(x2)^2) * root((x1)^2+(x2)^2 + (y1)^2+(y2)^2) / ((y1)^2+(y2)^2) * root(1 - (x1)^2-(x2)^2 / ((x1)^2+(x2)^2 + (y1)^2+(y2)^2)) (using the property of square roots)

= 2x1y1 / root((x1)^2+(x2)^2 + (y1)^2+(y2)^2) * root(1 - (x1)^2-(x2)^2 / ((x1)^2+(x2)^2