The Arithmetic-Geometric Mean Inequality Problem

[hsong9]

1. The problem statement, all variables and given/known data
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.


2. Relevant equations



3. 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..?

[Dick]

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.

[hsong9]

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).

[Dick]

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.

[hsong9]

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?

[Dick]

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.

[hsong9]

Thanks!!