Proof of this inequality

[#1procrastinator]

How does the result



prove the inequality for positive numbers? Doesn't that say that b must be equal to a or less?

[xZero]

if and only if

[dc302]

How does the result



prove the inequality for positive numbers? Doesn't that say that b must be equal to a or less?

You're right, the a>b is not necessary in proving the first result, as the first result is true for any non-negative a and b.

[Natters]

a wekk from the spesh exam and i truly dont get these kinds of things, like i can see it fine as in the average of a and b will the the root of ab if a=b and whatnot but i cant write what xzero just did... fuck that, where did you get the... how the... lol

[dc302]

a wekk from the spesh exam and i truly dont get these kinds of things, like i can see it fine as in the average of a and b will the the root of ab if a=b and whatnot but i cant write what xzero just did... fuck that, where did you get the... how the... lol

If you want to be able to think of that solution, it's easy. Simply start from the proposed inequality, then use it to work backwards to where you should start. For example, try moving the 2 across to the LHS and then squaring both sides, and go from there. Once you get to an obvious result, which is that (a-b)^2 > 0, then you can use that as your starting point and simply reverse your working. Just make sure you don't make any assumptions that may not be true conversely though.

[xZero]

actually i learnt it in 1 of my maths unit, we did it as an exercise to proof that

and dw, when i saw it the first time i was like wtf how the hell do you proof that

[Natters]

yeah but when you saw it for the first time it wasnt 5 days from the exam

[abeybaby]

arithmetic mean is always greater than or equal to the geometric mean.

[#1procrastinator]

How does the result



prove the inequality for positive numbers? Doesn't that say that b must be equal to a or less?

You're right, the a>b is not necessary in proving the first result, as the first result is true for any non-negative a and b.

I realised when you square root , you get OR

so if working backwards from the final result to get the inequality (like xZero did), then the inequality is true for the final result? (a > b, b < a, a = b)

[dc302]

I'm not sure what you mean. The inequality does not depend on the relative positions of a and b (as long as they're both positive). This can be seen if you simply swap a and b, and you will get the same inequality.

[#1procrastinator]

^ Yeah, that's why I started this thread. The initial result I got didn't make sense but I realised in the last post that when you square root it, you take both cases which then covers a < b, a = b, a > b, so it doesn't matter whether one is larger than the other like you said. Basically, that's what I wanted to prove

[xZero]

inequality holds as long as a doesn't equal to b, if a=b then equality holds. does that clear things up?

[#1procrastinator]

so that's basically what it says? when you prove something like this, then the final result are the numbers that the equation/inequality is true for?

[TrueTears]

very famous inequality, known as the AM-GM inequality, http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means

various methods of proof, my 2 faves are the geometric proof (still remember back in the days when kamil mentioned it in Re: TT's Maths Thread ) and a proof done by Zeitz in Art and Craft of Problem Solving (he used the rearrangement inequality http://en.wikipedia.org/wiki/Rearrangement_inequality)!