simple inequality in Rudin In his Functional Analysis book (2nd ed) Rudin asserts on page 203 that 
$$
(1+|y|)^{2r} < (2n+2)^r (1 + y_1^{2r}+...+y_n^{2r}) 
$$
where $r$ is a positive integer and $|y|$ is the Euclidean norm of the real vector $y = (y_1,...,y_n)$.  Can anyone see how to prove this?
 A: Recall the power mean inequality, which, in particular, tells you that
$$ \frac{\sum_{i=1}^{m} x_i}{m} \le \sqrt[s]{\frac{\sum_{i=1}^{m} x_i^s}{m}},$$
for any $s \ge 1$ and any non-negative $x_i$. Rudin's inequality is a double application of this. First apply it with $m=2, s=2, x_1=1, x_2 =|y|$:
$$(1+|y|)^2 \le 2(1+\sum_{i} y_i^2).$$
Then apply it again, this time with $m=n+1$, $s=r$, $x_i = y_i^2$ for $1 \le i \le n$ and $x_{n+1}=1$. This shall give you:
$$(1+\sum_{i} y_i^2)^r \le (n+1)^{r-1} (1 + y_1^{2r}+...+y_n^{2r}).$$
Now combine both inequalities.
A: By the way, the constant is not sharp (I guess that it is not important for Rudin's application). For finding the sharp constant, denote $c=|y|$ and note that by power mean inequality (cf. Ofir's answer) we have $\sum y_i^{2r}\geqslant n^{1-r}(\sum y_i^2)^r=n^{1-r}c^r$, with equality when all $y_i^2$ are equal. Next, we need to maximize $(1+c)^{2r}/(1+n^{1-r}c^{2r})$. Taking the logarithmic derivative of this expression, which equals $2r/(1+c)-2rn^{1-r}c^{2r-1}/(1+n^{1-r}c^{2r})$ we see that the minimum is attained for $c=n^{(r-1)/(2r-1)}$ and equals $(1+c)^{2r-1}=(1+n^{(r-1)/(2r-1)})^{2r-1}$. This is of course less than $(2n+2)^r$, since the exponent $n^{(r-1)/(2r-1)}<\sqrt{n}$ and $(1+\sqrt{n})^2\leqslant 2n+2$.
