Question: 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!

Answer: 
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 $n$.
**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 $x = c\_i^p$.

$$
\sqrt[p]{\frac{c\_1^p+\cdots +c\_n^p}{n}}
\ge \frac{c\_1+\cdots +c\_n}{n}\\
\sqrt[p]{{c\_1^p+\cdots +c\_n^p}}
\ge \frac{c\_1+\cdots +c\_n}{n^{1-1/p}}
$$