MATH HELP!

[GoldenBear]

ARG, I've been looking at this problem for 30 minutes now and still can't figure it out!!

Prove that..

sqrt(ab) <= (a+b)/2, and a<=b

 

[GoldenBear]

What I've done so far is...

sqrt(ab) <= (a+b)/2
2sqrt(ab) <= (a+b) - multiply each side by 2
4ab <= a^2+2ab+b^2 - square each side
2ab <= a^2+b^2 - subtract each side by 2ab

Now I can't go on any futher..

 

[lucidguy]

Dude, I get the feeling that we are helping you fill out your final take home math exam.

This is the 2nd or 3rd thing you posted in the past 24 hours? What is with that?

 

[GoldenBear]

<< Dude, I get the feeling that we are helping you fill out your final take home math exam.

This is the 2nd or 3rd thing you posted in the past 24 hours? What is with that? >>

Dude, I get the feeling you can't help me. Why waste your time here?

And uh, who has a final exam on June 24th?

 

[josphstalinator]

0 <= a^2 -2ab +b^2
0 <= (a-b)^2

done

 

[GoldenBear]

<< 0 <= a^2 -2ab +b^2
0 <= (a-b)^2

done >>

Eek, that's amazing.

Whoo, I can now sleep

 

[BruinEd03]

based on what u've got so far

0 <= a^2 - 2ab + b^2 (just shift that 2ab to the other side)
0 <= (a-b)^2 (factoring)
since a square of something is always greater or equal to 0, then a and b can be any number...hope that helps =)

-Ed

 

[BruinEd03]

bahhhh!! someone beat me to it =P

-Ed