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sumpm1

Member

Registered: 2007-03-05

Posts: 42

Real Analysis Help

Hello, we are covering absolute values, intervals and inequalities in my Real Analysis class. Specifically, we have recently covered the Geometric Sum, as well as the Arithmetic Mean/Geometric Mean Inequality that are stated below.

Geometric Sum: If

and

are real numbers, then

Arithmetic Mean/Geometric Mean Inequality: Let

be a positive integer. If

are nonnegative numbers, then

. Equality occurs if and only if

.

Problems:

1. Let

and

be fixed positive numbers. Find the value of

that minimizes the given expression and determine the minimum value of the expression.

a.

b.

c.

2. Let

be a positive integer and let

be positive numbers. The harmonic mean of these numbers is the reciprocal of the arithmetic mean of the reciprocals of the numbers. Prove that the harmonic mean of a set of positive numbers is less than or equal to the geometric mean. When does equality occur?

I am unclear of the method for finding the maximum or minimum of an expression. Ant help is appreciated.

Thanks

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Real Analysis Help

The most common way to find minima would be to differentiate and equate that expression to 0, which seems a bit weird since the stuff you're given has nothing to do with calculus.
I can't think of any way to apply geometric sequences to this question though.

---

Why did the vector cross the road?
It wanted to be normal.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Real Analysis Help

@sumpm1

Q1.
For 1a,

So the minimum value of

is

, which is attained if and only if

.

Q2
Hint: Apply the AM–GM inequality to

.

Last edited by JaneFairfax (2009-02-14 02:46:54)

Muggleton

Member

Registered: 2009-01-15

Posts: 65

Re: Real Analysis Help

1(b)

Hence the LHS is minimized

Use the same trick for 1(c), rewriting