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, and$a>1 = g$. 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 $x_{1}x_{2}\ldots x_{n}\neq 1$