Let $n \in \mathbb N$ and $x,y \in \mathbb R^n~;~x \ne y$. Let $p \in (1, \infty)$. Then, the function $f$ defined on $[0,1]$ to be $t \rightarrow \sum \limits_{i=1}^{n}|tx_i+(1-t)y_i|^p~$ attains its maximum value at either $0$ or $1$ and nowhere in between.
Proof Attempt:
$\Phi(t) = \sum \limits_{1}^{n} |tx_i+(1-t)y_i|^p = \sum \limits_{1}^{n} [a_i (tx_i+(1-t)y_i) ]^p $
where $a_i=1$ if $(tx_i+(1-t)y_i)$ is positive and $-1$ if negative.
$f$ is a continuous function defined on a compact set and thus it must attain its bounds. The above relation :
$\implies \Phi'(t) = \sum \limits_{1}^{n} p [a_i (tx_i+(1-t)y_i) ]^{p-1} \cdot a_i (x_i-y_i)$
$\implies \Phi''(t) = \sum \limits_{1}^{n} p \cdot (p-1)~ | ~(tx_i+(1-t)y_i) ~|^{p-2} \cdot (x_i-y_i)^2$
It's clear that $\Phi''(t) > 0 \implies \Phi'(t)$ is an increasing function.
If we prove that either $(i)~~\Phi'(0) > 0,$ then $\Phi'(t) > 0 ~\forall~ t$ which will imply that $\Phi$ is an increasing function and hence, we can obtain that $\Phi$ obtains its greatest value at $t =1 $. or we prove that $~(ii)~\Phi'(0) \rightarrow -\infty$ and $- \infty < \Phi'(1) < 0$, then $\Phi$ attains its greatest value at $t=0$.
$\Phi'(0) = p \sum \limits_{1}^{n} a_i^p y_i^{p-1} (x_i-y_i) $
How do I move forward from here?
Since you showed that $\phi'(t)$ is increasing, we have that $\phi$ is a convex function. Thus, we apply Jenson's inequality for $t \in [0,1] $: $$(1-t)\phi(0)+t\phi(1)> \phi((1-t)*0+t*1)=\phi(t)$$ If $a=\max(\phi(0),\phi(1))$, then we have $a>\phi(t) \forall t \in (0,1)$. Thus, $a$ is the maximum value of $\phi$ which is attained at $t=0$ or $t=1$