1. ## prove inequality

x is less then or equal to 0
y is less then or equal to 0
then prove (xy)^1/2 less then or equal to x+y/2

I was wondering if someone could tell me where to start It would be greatly appreiated i forgot this on my last post. Thank you very much. I am going crazy.

2. This is the arithmetic-mean geometric mean inequality.

Let x>=0 and y>=0.

Then,
0<=(x-y)^2 because a square is always non-negative.
Expand,
0<=x^2-2xy+y^2
4xy<=x^2+2xy+y^2=(x+y)^2
Take square roots since both are non-negative,
sqrt(4xy)<=sqrt(x+y)^2
2sqrt(xy)<=|x+y|=x+y
Divide by 2,
sqrt(xy)<=(x+y)/2

3. Originally Posted by schinb64
x is less then or equal to 0
y is less then or equal to 0
then prove (xy)^1/2 less then or equal to x+y/2

I was wondering if someone could tell me where to start It would be greatly appreiated i forgot this on my last post. Thank you very much. I am going crazy.
as (xy)^(1/2) is positive, and (x+y)/2 is negative this is not true
the way you have asked the question.

ImPerfectHacker has answered this but for:

x >= 0
y >= 0,

which is almost certainly what you were really asked to do.

RonL

4. ## Thank you for the responses you guys are great. The help is greatly appreciated

Thank you for the help again. You guys are great. You have helped me so much thank you again
Originally Posted by ThePerfectHacker
This is the arithmetic-mean geometric mean inequality.

Let x>=0 and y>=0.

Then,
0<=(x-y)^2 because a square is always non-negative.
Expand,
0<=x^2-2xy+y^2