The Arithmetic-Geometric Mean Inequality Problem

In summary, we need to verify that f(x) = log(1+ex) is convex and then use this fact to show that (1+x1)1/n(1+x2)1/n..(1+xn)1/n >= 1 + (x1x2...xn)1/n for positive real numbers x1, x2, ..., xn. To do this, we can rewrite the inequality in terms of f(x) and use Jensen's inequality, which states that for a convex function f(x), f(∑aix(i)) <= ∑aif((i)) when a1,...,ak are nonnegative numbers with sum 1 and x(1),...,x(k) are points in
  • #1
hsong9
80
1

Homework Statement


verify that f(x) = log(1+ex) is convex and use this to show that
(1+x1)1/n(1+x2)1/n..(1+xn)1/n >= 1 + (x1x2...xn)1/n
where x1,x2,...xn are positive real numbers.


Homework Equations





The Attempt at a Solution


I know f(x) is convex b/c f''(x) is always positive.
and I can guess the above ineqality is true, but
I do not konw how I use the f(x) to prove the inequality.
Can log (1+e^x) imply to (1+x1)1/n...(1+xn)1/n..?
 
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  • #2
Start by writing xi=e^(log(xi) and try to express both sides of the inequality in terms of the function f(x). Taking the log of both sides would be a good start.
 
  • #3
Thanks,
so.. f(x) = (1+ex)

(1+x1)1/n...(1+xn)1/n =
(1+elnxn)1/n ==> f(x)
Thus,
Σ((1/n)log(1+elnxi)) = Sigma (f(x)*(1/n))

Also,
log (1+ (ex1...xn)1/n
= log (1 + eln Σxi(1/n))
Therefore,

Σ(f(x)*(1/n)) >= f(Σ(1/n)*xi) as required. right?

I am not sure
log (1 + eln Σ(1/n)*xi) implies to f(Σ(1/n)*xi).
 
  • #4
I'm not sure that is quite correct, since you seem to have dropped the log's at the end. But one side should be the mean of f(log(xi)) and other side should be f(mean log(xi)) where 'mean' is the arithmetic mean. If f is convex, what's the relation between those two? Hint: if f is convex then f((a+b)/2)<=(f(a)+f(b))/2. That can be generalized to more than just two values a and b.
 
Last edited:
  • #5
Sorry, my writing was confused.

I konw Jensen’s inequality.
Suppose that f(x) is a convex function defined on a convex subset C of Rn.
If a1,...,ak are nonnegative numbers with sum 1 and if x(1),...,x(k) are points of C, then
f(∑aix(i)) <= ∑aif((i)).

f(x) = log(1+ex)
(1+x1)1/n...(1+xn)1/n >= 1 + (x1..xn)1/n

(1+x1)1/n...(1+xn)1/n -->
(1/n)log(1+elog(x1))+...(1/n)log(1+elog(xn)) = ∑(1/n)log(1+elog(xi))

1 + (x1..xn)1/n -->
log(1 + elog(x1..xn)1/n) = log(1 + e(1/n)log(x1)+..(1/n)log(xn)) = log(1 + e∑(1/n)log(xi))

Therefore,
∑(1/n)log(1+elog(xi))>= log(1 + e∑(1/n)log(xi))
This inequality is equivalent to the Theorem as required inequality.

Right?
 
  • #6
hsong9 said:
Sorry, my writing was confused.

I konw Jensen’s inequality.
Suppose that f(x) is a convex function defined on a convex subset C of Rn.
If a1,...,ak are nonnegative numbers with sum 1 and if x(1),...,x(k) are points of C, then
f(∑aix(i)) <= ∑aif((i)).

f(x) = log(1+ex)
(1+x1)1/n...(1+xn)1/n >= 1 + (x1..xn)1/n

(1+x1)1/n...(1+xn)1/n -->
(1/n)log(1+elog(x1))+...(1/n)log(1+elog(xn)) = ∑(1/n)log(1+elog(xi))

1 + (x1..xn)1/n -->
log(1 + elog(x1..xn)1/n) = log(1 + e(1/n)log(x1)+..(1/n)log(xn)) = log(1 + e∑(1/n)log(xi))

Therefore,
∑(1/n)log(1+elog(xi))>= log(1 + e∑(1/n)log(xi))
This inequality is equivalent to the Theorem as required inequality.

Right?

That's it.
 
  • #7
Thanks!
 

What is the Arithmetic-Geometric Mean Inequality Problem?

The Arithmetic-Geometric Mean Inequality Problem is a mathematical conjecture that states that for any two positive real numbers, their arithmetic mean (the average) is always greater than or equal to their geometric mean (the square root of their product).

Who came up with the Arithmetic-Geometric Mean Inequality Problem?

The Arithmetic-Geometric Mean Inequality Problem was first stated by French mathematician Joseph Louis Lagrange in the late 18th century.

Is the Arithmetic-Geometric Mean Inequality Problem proven to be true?

As of now, the Arithmetic-Geometric Mean Inequality Problem remains an open problem in mathematics. It has been proven to be true for certain special cases, but a general proof has not yet been found.

Why is the Arithmetic-Geometric Mean Inequality Problem important?

The Arithmetic-Geometric Mean Inequality Problem has many applications in different areas of mathematics and science, such as probability, optimization, and engineering. It also serves as an important tool in solving other mathematical problems.

What progress has been made towards solving the Arithmetic-Geometric Mean Inequality Problem?

Over the years, many mathematicians have attempted to prove the Arithmetic-Geometric Mean Inequality Problem, but none have been successful so far. Some have made progress in proving special cases or related conjectures, but a complete proof remains elusive.

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