Prove $ \frac {x_1 + \cdots + x_n}{n} \ge \sqrt[n]{x_1 x_2 \cdots x_n}$ I am studying computer science in the first term.
I have to proof the following inequality:

$$ \frac {x_1 + \cdots+ x_n}{n} \ge \sqrt[n]{x_1 x_2 \cdots x_n}$$

$x$ can be any positive real Number: 
$ x \in \mathbb{R},x \gt0 $
I try to bring it in a form that is like Bernoulli's Inequality, but then realized that this is pretty much nonsense.
So now I am having a hard time to prove it through induction.
Beginning of the induction is clear: $n \rightarrow 1$
$ \large\frac {x_1}{1} \ge x_1$
Next Step: $n \rightarrow n+1$
For the start, lets write the left side of the inequality as
$$x_a(n) = \frac{1}{n} \sum_{k=1}^n x_k$$ 
$$x_g(n) =  \prod_{k=1}^n x_k^\frac{1}{n}$$
So now we have to prove:
$$x_a(n+1) = \frac{1}{n+1} \sum_{k=1}^{n+1} x_k$$ 
$$x_g(n+1) =  \prod_{k=1}^{n+1} x_k^\frac{1}{n+1}$$
$x_a(n+1) \ge x_g(n+1)$
Can you just give me a strategy on that?
Thanks.
 A: Hints: The standard way to solve this: Put $f(x) = x_1 x_2 ... x_n $. Let 
$$ D = \{ x \in \mathbb{R}^n: x_1 + ... + x_n = n , x_i \geq 0 \}$$
Now, use LAgrange multipliers to find the max of $f$ on the set $D$. 
Conclude.
A: Many ways, here is another one:
Note that the inequality is homogeneous, (we can scale all variables by any positive value and the inequality remains unchanged). So WLOG we can set $x_1 x_2\dots x_n = 1$.  Then we need to show only that $x_1+x_2 + \dots + x_n \ge n$ under this condition.
Consider the function $f(t) = t-1-\log t$.  Note that our inequality is the same as $f(x_1)+f(x_2)+\dots + f(x_n)\ge 0$, so it is enough to show $f(t)\ge 0$ for positive $t$.  This is easily done using one variable calculus or if you know the series expansion of $\log (1+t)$ or $e^t$.
A: An elementary way:
 By squaring both relations on the left side of the implication, it's easy to see this holds:
$$x+y=x'+y', |x-y|\ge|x'-y'|\implies x'y'\ge xy.$$
Now set $a=(x_1+\ldots+x_n)/n$. Find $x_i\ne a$ so that $|x_i-a|$ is the smallest possible. If $x_i>a$, take any $x_j<a$, then by using the above inequality you can replace $x_i$ by $a$ and $x_j$ by $x_j{+}(x_i{-}a)\le a$; similarly for $x_i<a$. Repeat until all $x_i = a$.
A: I think the easiest way to prove this is to use the concavity of ln:
$(ln(x))" = -\frac{1}{x^2} < 0$ => ln concave
Then you know that if: $\sum_{k=1}^n a_k = 1$ : $ln(\sum_{k=1}^n a_k*x_k) \geq \sum_{k=1}^n a_k*ln(x_k)$
Now set: $a_k =\frac{1}{n}$ for $k=1..n$, and see what you get, knowing that: $ a*ln(x)=ln(x^a)$ and $\sum_{k=1}^n ln(x_k) = ln(x_1*x_2*...*x_n)  $
A: Another Solution:
First, a result: for any $x,y$ non-negative reals, $\alpha + \beta = 1 $, then
$$ x^{\alpha}y^{\beta} \leq \alpha x + \beta y $$
To show this, put $f(t) = (1 - \beta) + \beta t - t^{\beta} $. Show this function decreases on $[0,1]$ and then replace $t$ with $\frac{y}{x}$. . Now, as for your problem, we use induction on $n$. The base case is Young, and now let us suppose result holds for $n$, we show it holds for the case $n+1$. Indeed,
$$x_1^{\alpha_1}...x_{n+1}^{\alpha_{n+1}} = [ x_1^{ \frac{\alpha_1}{\alpha_1 + ... + \alpha_n}}...x_n^{ \frac{\alpha_n}{\alpha_1 + ... + \alpha_n}} ]^{\alpha_1 + ... + \alpha_n}x_{n+1}^{\alpha_{n+1}} \leq (\alpha_1 + ... + \alpha_n )x_1^{ \frac{\alpha_1}{\alpha_1 + ... + \alpha_n}}...x_n^{ \frac{\alpha_n}{\alpha_1 + ... + \alpha_n}} + \alpha_{n+1}x^{n+1}$$
$$ \leq (\alpha_1 + .... + \alpha_n)[ \frac{\alpha_1}{\alpha_1 + ... + \alpha_n}x_1 + ... + \frac{\alpha_n}{\alpha_1 + ... + \alpha_n}x_n] + x_{n+1}\alpha_{n+1}$$
By induction hypothesis. But, the the above is obviously equal to
$$ = \alpha_1 x_1 + .... + \alpha_{n+1}x_{n+1} $$
Hence, your result follows by math induction.
