How to prove the inequality involving a sum of a sequence divided by $\sqrt{n}$ I'm trying to prove # 19 on page 41 of Bartle and Sherbert's Real Anaylsis.  Here is the problem:
$\forall k$ let $c_k>0$, show:
$$
\frac{c_1+c_2+...+c_n}{\sqrt{n}}\leq (c_1^2+c_2^2+...+c_n^2)^\frac{1}{2}
$$
I'm having trouble showing this with induction, and I am hopeful that someone might give a hint to point me in a good direction.  Thanks!
 A: We can change this problem to be: $$(c_1 + c_2 + \cdots + c_n)^2 \le n (c_1^2 + c_2^2 + \cdots + c_n^2)$$
Then we can establish a quick lemma: $$x^2 + y^2 \ge xy + yx.$$
This is established by noticing: $(x-y)^2 \ge 0$ and so $x^2 + y^2 \ge 2xy = xy + yx$.
Now if we consider the case when $n=2$ we see that
$$(c_1 + c_2)^2 = c_1^2 + c_1 c_2 + c_2 c_1 + c_2^2 \le c_1^2 + c_1^2 + c_2^2 + c_2^2 = 2(c_1^2 + c_2^2).$$
The same reasoning can be carried out for arbitrary $n$.
A: Hint: use the convexity of $x\to x^2$. 
details:
$$\left(
\frac{x_1+\cdots +x_n}n
\right)^2 \le
\frac{x_1^2+\cdots +x_n^2}n\\
\frac{x_1+\cdots +x_n}n
 \le
\frac{\sqrt{x_1^2+\cdots +x_n^2}}{\sqrt{n}}\\
\frac{x_1+\cdots +x_n}{\sqrt{n}}
 \le
{\sqrt{x_1^2+\cdots +x_n^2}}
$$

alternatively, you can use concavity of $x\to x^{1/p}$,  with $p\ge 1$ ; in you example $p=2$:
$$
\sqrt[p]{\frac{x_1+\cdots +x_n}{n}}
\ge \frac{\sqrt[p]{x_1}+\cdots +\sqrt[p]{x_n}}{n}
$$
Take $x = c_i^p$.
$$
\sqrt[p]{\frac{c_1^p+\cdots +c_n^p}{n}}
\ge \frac{c_1+\cdots +c_n}{n}\\
\sqrt[p]{{c_1^p+\cdots +c_n^p}}
\ge \frac{c_1+\cdots +c_n}{n^{1-1/p}}
$$
