prove the inequalities: $|x|_q \leq |x|_p \leq N^{^1/p-^1/q}\cdot |x|_q, x \in \mathbb{R}^{N}$ and $||x||_{l^q} \leq ||x||_{l^p}$ for $x\in l^p$ I want to show two inequalities in normed spaces:
1.)  $$|x|_q \leq |x|_p \leq N^{^1/p-^1/q} \cdot |x|_q, \hspace{3mm} x \in \mathbb{R}^{N}$$ and
2.)$$||x||_{l^q} \leq ||x||_{l^p},\hspace{5mm} x\in l^p$$
with $$1\leq p\leq q \leq \infty.$$
Edit: I screwed up the order of p and q in the inequality at first.
 A: *

*Let us prove that $\|x\|_q \leq \|x\|_p$ for every $x\in \ell^p$ (the proof is the same in $\mathbb{R}^n$).


Clearly it is enough to consider the case $x\neq 0$.
As you have already observed, the proof is not difficult for a normalized vector.
Hence, let us define
$$
y := \frac{x}{\|x\|_p}
$$
so that $\|y\|_p = 1$. In particular, $|y_k|\leq 1$ for every $k$, so that $|y_k|^q \leq |y_k|^p$ and
$$
\|y\|_q = (\sum |y_k|^q)^{1/q} \leq (\sum |y_k|^p)^{1/q} = \|y\|_p^{p/q} = 1.
$$
Hence, by the very definition of $y$, it follows that $\|x\|_q \leq \|x\|_p$.


*Let us prove that $|x|_p \leq N^{1/p - 1/q} |x|_q$ for every $x\in \mathbb{R}^N$.
Let us consider the exponent $\alpha := q/p > 1$ and its conjugate exponent $\alpha' = q/(q-p)$. By Holder's inequality we have that
$$
\sum |x_k|^p \leq \left[ \sum (|x_k|^p)^\alpha \right]^{1/\alpha} N^{1/\alpha'}
$$
i.e.
$$
|x|_p^p \leq |x|_q^p N^{(q-p)/q}
\quad
\Longrightarrow
\quad
|x|_p \leq |x|_q N^{1/p - 1/q}.
$$

A: 1.) I recall the Power Mean inequality. Let $x_1,\dots, x_n$ be positive real numbers. The power mean of order $r$ is defined by
$M(x_1,\dots, x_n)(0) =\sqrt[n]{x_1\cdots x_n}$, $M(x_1,\dots,x_n)(r) =\left(\frac{x_1^r+\dots+x_n^r}{n}\right)^{\frac 1r}$ $(r\ne 0)$.
Then the function $M(x_1,\dots x_n):\Bbb R\to\Bbb R$ is continuous and non-decreasing. This claim should follow from the convexity of $x\ln x$ or the convexity of $x^\lambda$ $(\lambda\ge 1)$ and Jensen's inequality. When $r\ge 0$ then a similar claim holds for non-negative $x_1,\dots, x_n$.
Now let $x=(x_1,\dots, x_N)\in\Bbb R^N$. Put $x^*=(|x_1|,\dots, |x_N|)$. 
Then $|x|_q=M(x^*)N^{-1/q}$, $|x|_p=M(x^*)N^{-1/p}$. 
So $|x|_q N^{1/q}\le |x|_p N^{1/p}$.
