If the product of n numbers is 1, their sum is >= n

[matticus]

i'm reading shilov's "real and complex analysis". there is a problem to prove that for real positive x1,x2,...xn, if x1*x2*...*xn=1, then x1+x2+...xn >= n. I proved this for the case n = 2. it says use induction on this case to prove it for the nth case. but i just don't see it.

for n=2:

(a+1)^2 >= 0
a^2 + 2a + 1 >= 0
a^2 + 1 >= 2a
a + 1/a >= 2

 

[Kummer]

i'm reading shilov's "real and complex analysis". there is a problem to prove that for real positive x1,x2,...xn, if x1*x2*...*xn=1, then x1+x2+...xn >= n. I proved this for the case n = 2. it says use induction on this case to prove it for the nth case. but i just don't see it.

Use the AM-GM inequality.

$$\frac{x_1+x_2+...+x_n}{n} \geq \sqrt[n]{x_1x_2...x_n} = 1$$

So,
$$x_1+...+x_n \geq n$$

 

[matticus]

right. actually it's a later problem to prove that inequality. but that's not really what I'm looking for. he asks as one problem to prove that a + 1/a >= 2. then he says use induction to prove it for the general case. so what i really want is an inductive proof.

 

[Klaus_Hoffmann]

Use the AM-GM inequality.

$$\frac{x_1+x_2+...+x_n}{n} \geq \sqrt[n]{x_1x_2...x_n} = 1$$

So,
$$x_1+...+x_n \geq n$$

hey, this AM-GM equality gives me a problem, let's suppose we would like to generalize it to an infinite dimensional space, the AM is:

$$\int_{a}^{b}dx (b-a)^{-1} f(x)$$

however what would be the analogue of the GM ?

 

[Kummer]

however what would be the analogue of the GM ?

We need the restriction
$$f(x) >0$$

Then, perhaps this is a reasonable generalization,
$$\exp \left(\int_a^b \ln f(x) dx\right)$$

 

[StatusX]

We need the restriction
$$f(x) >0$$

Then, perhaps this is a reasonable generalization,
$$\exp \left(\int_a^b \ln f(x) dx\right)$$

That's exactly right, and both inequalities can be proven essentially the same way using http://en.wikipedia.org/wiki/Jensen%27s_inequality" , as shown in chapter 3 of Rudin's analysis book.