Proof of Simple Inequality for Positive Real Values x_1,x_2,...,x_n

[sat]

Let x_{1},x_{2},\ldots,x_{n} be real and positive. Show that
g\equiv \sqrt[n]{x_{1}x_{2}\ldots x_{n}} < \frac{x_{1}+x_{2}+\ldots+x_{n}}{n} \equiv a
except when x_{1}=x_{2}=\ldots = x_{n} in which case a=g.

We are given the results (from previous exercises) that
\forall x>0 : x+\frac{1}{x} \geq 2
and if x_{1}x_{2}\ldots x_{n}=1, then
x_{1}+x_{2}+\ldots+x_{n}\geq n
and this is equality if and only if x_{1}=x_{2}=\ldots=x_{n}=1.

Clearly if x_{1}x_{2}\ldots x_{n}=1, then the last result applies directly, anda>1 = g. When x_{1}x_{2}\ldots x_{n}\neq 1 things are more difficult.

I can show that, for x_{1}x_{2}\ldots x_{n}<1, a>g^{n}=x_{1}x_{2}\ldots x_{n}, but g>g^{n} which doesn't help. To do that I used the first `given' result to derive that
x_{1}+x_{2}+\ldots + x_{n} \geq 2n - \frac{x_{1}+x_{2}+\ldots + x_{n}}{x_{1}x_{2}\ldots x_{n}}
and took it from there.

I'm also not making much progress with the case x_{1}x_{2}\ldots x_{n}>1. I have tried doing things like `forcing' the expressions to equal unity by introducing x_{1}^{-1} etc and increasing n accordingly but the problem is then to show that a is greater than the new mean including these reciprocals and that the geometric mean is less than this.

Any ideas would be apprectiated. (It's probably something very simple which I'm overlooking.)

The problems are from Shilov "Introductory Real and Complex Analysis", p24 if anyone has that.

 

[Popey]

Hi!

if x_{1}x_{2}\ldots x_{n}=1 (1), then
x_{1}+x_{2}+\ldots+x_{n}\geq n
and this is equality if and only if x_{1}=x_{2}=\ldots=x_{n}=1.

Set P=x_{1}x_{2}\ldots x_{n} then
g\equiv \sqrt[n]{x_{1}x_{2}\ldots x_{n}} =\sqrt[n]{P}

For the real numbers w_{i}=x_{i}/g we have

w_{1}w_{2}\ldots w_{n}=(x_{1}/g)(x_{2}/g)\ldots (x_{n}/g)=P/g^n=1

So, (1) holds for w_i. Then

w_{1}+w_{2}+\ldots+w_{n}\geq n (this is an equality if and only if w_{i}=1 <=> x_{i}=g)


\frac{x_{1}+x_{2}+\ldots+x_{n}}{g}\geq n

\frac{x_{1}+x_{2}+\ldots+x_{n}}{n}\geq g

That is,
a\geq g and this is an equality if and only if x_{i}=g=\sqrt[n]{P}, \forall i

 

[sat]

Thanks. That seems to solve the problem of making it work for x_{1}x_{2}\ldots x_{n}\neq 1