Fundamental Optimization question consisting of two parts. A) Find all extrema of $$f(x)=\sum_{k=1}^{n} x_{k}^{2} $$ subject to the constraint $\sum_{k=1}^{n}\vert x_k\vert^p=1$
B) prove that $$\frac{1}{n^{(2-p)/(2p)}}(\sum \vert x_k\vert^p)^{(1/p)}\le (\sum x_k^2)^{1/p} \le (\sum \vert x_k\vert^p)^{(1/p)}$$
for $k=1,\dots ,n$ 
$\forall x_1, \dots x_n \in \Bbb R \ \& \  n\in \Bbb N \ \& \ 1\le p\le 2. $ 

There exists its solution. But this is too complicated,  not simple and clear. So I dont understand it. Please help me how should I solve this in a clear way? 
 A: OK, here is how I would do it by using Lagrange multipliers.
A. Since $f(x)$ does no change if one changes the sign of some coordinates of $x$, it is enough to find the extrema of $f(x)$ subject to the constraints $x_1,\dots ,x_n\geq 0$ and $\sum_1^n x_i^p=1$. Moreover, we can of course assume that $p\neq 2$, because the exercise is trivial for $p=2$
First note that 
$$\Sigma=\left\{ (x_1,\dots ,x_n);\; x_i\geq 0\;{\rm and}\; \sum_{i=1}^nx_i^p=1\right\} $$
is a compact subset of $\mathbb R^n$ (it is closed an bounded), so your continuous function $f$ does achieve a maximum and a minimum.
Assume that $f$ attains an extremum at some point $a=(a_1,\dots ,a_n)\in\Sigma$. By permuting the coordinates, we may assume that we have $a_1,\dots ,a_s>0$ for some $s\in\{ 1,\dots ,n\}$ and $a_{s+1}=\dots =a_n=0$ (if $s<n$). If we denote by $\Omega_s$ the open set $\{ x_1>0,\dots ,x_s>0\}$ in $\mathbb R^s$, then we can try to apply the Lagrange multipliers theorem in $\Omega_s$ to the function $f_s(x_1,\dots ,x_s)=\sum_1^s x_i^2$ and the constraint $g(x_1,\dots ,x_s)=0$, where $g(x_1,\dots ,x_s)=\sum_1^s x_i^p-1$. The function $f_s$ and the constraint function $g$ are indeed $\mathcal C^1$ in $\Omega_s$, and $f_s$ has an extremum at $(a_1,\dots ,a_s)$ in the open set $\Omega_s$ with respect to the constraint "$g=0$". Since $\nabla g(a_1,\dots ,a_s)=(pa_1^{p-1},\dots ,pa_{s}^{p-1})\neq 0$, the Lagrange theorem can indeed be applied; and this gives some "multiplier" $\lambda$ such that $\nabla f_s(a_1,\dots ,a_s)=\lambda \nabla g(a_1,\dots ,a_s)$. In other words, we have $2a_i=\lambda p a_i^{p-1}$ for all $i\in\{ 1,\dots ,s\}$. Since $p\neq 2$, this gives $a_i=(\lambda p/2)^{1/p-2}$, so $a_1,\dots ,a_s$ are all the same. But $\sum_1^s a_i^p=\sum_1^n a_i^p=1$, so this gives $a_1=\dots =a_s=s^{-1/p}$ and $a_{s+1}= \dots =a_n=0$ (if $s<n$). Thus, we get $f(a)=\sum_1^na_i^2=\sum_1^s a_i^2=s\times s^{-2/p}=s^{1-\frac{2}p}$.
In summary, the extremal values of $f$ on $\Sigma$ are to be found amongst the $n$ numbers $s^{1-\frac2p}$, $1\leq s\leq n$. But $p<2$, so the exponent $1-\frac2p$ is $<0$. Hence, the maximal value is $1$ and the minimal value is $n^{1-\frac2p}$. 
The minimal value is attained at only one point of $\Sigma$, namely $a=(n^{-1/p},\dots ,n^{-1/p})$, and the minimal value is attained at the $n$ basic vectors $e_i=(0,\dots ,1,\dots ,0)$ (recall that we have used a permutation of coordinates in the beginning). As for the original question, then you get all maximizing and minimizing points from the ones we have found by changing arbitrarily the signs of some coordinates.
B. Part (A) says the following: whenever $u_1,\dots ,u_n\in\mathbb R$ satisfy $\sum_1^n \vert u_i\vert^p=1$, it follows that
$$n^{1-\frac2p}\leq \sum_{i=1}^n u_i^2\leq 1 \, ;$$
or, equivalently:
$$ \frac1{n^{\frac1p-\frac12}}\leq \left(\sum_{i=1}^n u_i^2\right)^{\frac12}\leq 1\, .$$
Now, if you take arbitrary $x_1,\dots ,x_n$ (not all $0$), then you can apply this to $u_i=\frac{x_i}{(\vert x_1\vert^p+\dots +\vert x_n\vert^p)^{1/p}}$, and this gives the required inequality:
$$ \frac1{n^{\frac1p-\frac12}}\left(\sum_{i=1}^n \vert x_i\vert^p\right)^{\frac1p}\leq \left(\sum_{i=1}^n x_i^2\right)^{\frac12}\leq \left(\sum_{i=1}^n \vert x_i\vert^p\right)^{\frac1p}\, .$$
