Proof questions

[Sovalovah]

For 6i how do i prove this statement?
Here is what i tried, which hasnt worked

 

[Harry_the_cat]

I think you are confusing pos and neg with even and odd.

 

[Dr.Peterson]

I think you are confusing pos and neg with even and odd.

And if the problem were about even and odd, you would be assuming that x and y are equal or differ by one. Next time you have a proof like that to do, let x = 2m and y = 2n, rather than both being 2n.

But that's just an idea to keep in mind for the future. What you've done is entirely wrong for the present problem.

 

[Zermelo]

You should investigate some proof techniques. The most simple ones are "direct proof" and "proof by contradiction".
For the direct proof, try to find an inequality that you know for sure holds true, and that is somehow connected to your problem (has a squareroot/square, two variables and a 2) and then play around with the inequality to try and get what you need to proove. (HINT: [MATH]x^2 \geq 0, \forall x \in R[/MATH])
For the proof by contradiction, suppose the opposite: [MATH]x,y>0,\ \ \sqrt{xy} > \frac{x+y}{2}[/MATH] and try to reach a contradiction.

 

[Steven G]

Another mistake you made is when you squared (4n+1). You just squared 4n and squared 1 and added the two. You think that (a+b)^2 = a^2 + b^2 which is not true. What does (a+b)^2 equal? What does (4n+1)^2 equal.

Why would you assume that the x and y are equal or differ by 1?????

 

[pka]

View attachment 24244
For 6i how do i prove this statement?

There is a straightforward way to prove this:
\(({\sqrt{x}-\sqrt{y})^2\ge 0}\\x-2\sqrt{xy}+y\ge 0\\x+y\ge 2\sqrt{xy}\\\dfrac{x+y}{2}\ge \sqrt{xy}\).