Inequalities (1 Viewer)

[Lainee]

(If you have a copy of Patel's Extension 2 book in front of you, it'll be easier cause what I'm referring to is Question 1 of the worked examples)

Prove (a+b)/2 >= sqrt.(ab) if a and b are positive real numbers.

It was pointed out to me that this basically means proving the arithmetic average >= the geometric average. To show what I mean, for example:

PROBLEM: Show (a+b+c+d)/4 >= 4th root. (abcd)
SOLUTION:
(a+b)/2 >= sqrt.(ab)
(a+b)/4 >= sqrt.(ab)/2 [1]

(c+d)/2 >= sqrt.(cd)
(c+d)/4 >= sqrt.(cd)/2 [2]

(assuming arithmetic average >= geometric average) -->
sqrt.(ab)/2 + sqrt.(cd)/2 >= sqrt.(sqrt.(ab)sqrt.(cd))
sqrt.(ab)/2 + sqrt.(cd)/2 >= 4th root. (abcd)

Because of [1] and [2] (I'm thinking, LHS of [1] and [2] are >= to RHS so of course if I sub the LHS in, the inequation direction above will still hold true)

.'. (a+b)/4 + (c+d)/4 >= 4th root. (abcd)
.'. (a+b+c+d)/4 >= 4th root. (abcd)


Okay, well you see my basic assumption here in order to prove this inequality is that the arithmetic average of a and b, and c and d >= their geometric average. So, CAN I ASSUME THIS without proving it? My parents were both schooled with an asian curriculum and they said it was perfectly acceptable over there to assume it. But will they accept it here? I've seen some questions where if you -can- assume it, it'll take you 2 mins to do it, but doing it a different way will take you heaps longer.

EDIT: I skipped a few explanative steps in the solution, but hopefully most would understand what I did in between.

 

[turtle_2468]

Hmm.
Usually the answer is no.
Especially in questions of the type
prove that (a+b+c)/3>=cube root(abc), which are just proving a specific case of the AM-GM.

However, the usualy method of proving it isn't too hard, and only takes marginally longer. (you prove it for 2, 4, 8, .. for all powers of 2 inductively, then substitute back... tell me if you want more of that one)

In practice HSC-wise, you usually only need to show it for 3 terms and 4 terms.

 

[Lainee]

Alright, I guess you better show me how to do it the usual method then. I've been assuming that all along until I looked in a standard textbook and got worried. Blah...

Can you show me how to do another question the accepted way then? I didn't think I could do this next one any other way then using AM>=GM - I'm obviously wrong.

The area of a triangle is given by the formula A= sqrt. (s(s-a)(s-b)(s-c)) where a, b, c are the lengths of the sides and s= (a+b+c)/2. Show that A <= (a^2 b^2 c^2)/4.

 

[turtle_2468]

The proof for 4 is relatively simple. In general, you know that (a+b)/2>=root(ab) and (c+d)/2>=root(cd). So (a+b+c+d)/2>=root(ab)+root(cd)
and so (a+b+c+d)/4>=(root(ab)+root(cd))/2
and apply the 2-variant AM-GM to the RHS again for the result.

What I mean is, you do use the AM-GM, but if you want to make sure u get full marks, prove it as well. In the above case, I proved it for 4 variables (a,b,c,d). To prove it for 3, just substitute d=(a+b+c)/3 and cancel out the power of (a+b+c) that occurs in common.

*Optional extra knowledge of how to prove 5,6,7,8,... variable AM-GM*:

As I did it for 4 above, getting 2 variables at a time then using the results in another one, you can do it for 8 variables. (a+b)/2>=root(ab), do the same for c+d, e+f, g+h. Then apply to root(ab) and root(cd), as well as root(ef) and root(gh). Then apply to 4th root(abcd) and 4th root(efgh). Hope that made sense...

 

[KeypadSDM]

The proof for AM >= GM isn't that hard, it's like 3 lines:

(Sqrt[A] - Sqrt)^2 >= 0
A + B >= 2Sqrt[AB]
(A + B)/2 >= Sqrt[AB]