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

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SUMMARY

The discussion centers on proving the inequality involving non-zero constants x1, x2, y1, and y2, specifically demonstrating that (2(x1)(y1))/(sqrt((x1)^2+(x2)^2)sqrt((y1)^2+(y2)^2)) ≤ (x1)^2/(x1)^2+(x2)^2 + (y1)^2/(y1)^2+(y2)^2. The user successfully simplifies the problem to the form 2xy ≤ x² + y², which is a direct application of the Arithmetic Mean-Geometric Mean inequality. The conclusion affirms that the method used is correct and aligns with established mathematical principles.

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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...
 
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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
 

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