Find max $x\cdot y$ where $x\in S= \{(x_1,\ldots, x_n)\in \mathbb{R}^n$ Let $S= \{(x_1,\ldots, x_n)\in \mathbb{R}^n$; $|x_1|^p+\ldots+|x_n|^p=1\}$, where $p>1$ is real(and fixed), consider a fixed $y\in\mathbb{R}^n$ and $T:\mathbb{R}^n\rightarrow\mathbb{R}$ such that $T(x) = x\cdot y$, where $x\cdot y = x_1y_1+\ldots+x_ny_n$.
I'm having a hard time to find $\max_{x\in S}\ T(x)$.
I already noticed a few things but its still really difficult to do something useful.
1) $\forall (x_1\ldots, x_n)\in S, |x_1|^p+\ldots+|x_n|^p\leq |x_1|+\ldots+|x_n|$;
2) Taking the norm $\Vert(x_1,\dots, x_n)\Vert=(|x_1|^p+\ldots+|x_n|^p)^{\frac{1}{p}}$ and the ball $B(0,1)$, with center $0\in\mathbb{R}^n$ and radius $1$, we have $S=\partial B$;
3) $\forall i=1\ldots n, T(e_i) = y_i$.
Also, I'm trying to avoid Holder's inequality. 

Define $\newcommand{\sgn}{\operatorname{sgn}}$
$$
F(x)=\sum_{k=1}^n|x_k|^p\tag{1}
$$
then
$$
\nabla F(x)=\left(p\sgn(x_k)|x_k|^{p-1}\right)_{k=1}^n\tag{2}
$$
You want to find a point on the surface where $\nabla F\,||\,y$. Therefore,
$$
x=\left(\sum_{k=1}^n|y_k|^{\frac{p}{p-1}}\right)^{-\frac{1}{p}}\left(\sgn(y_k)|y_k|^{\frac{1}{p-1}}\right)_{k=1}^n\tag{3}
$$
should be the point where $T(x)$ is the greatest. Computing $T(x)$ yields
$$
T(x)=\left(\sum_{k=1}^n|y_k|^{\frac{p}{p-1}}\right)^{\frac{p-1}{p}}\tag{4}
$$
