$\newcommand{\ep}{\epsilon}
\newcommand{\ga}{\gamma}
\newcommand{\la}{\lambda}
\newcommand{\Si}{\Sigma}
\newcommand{\R}{\mathbb{R}}
\newcommand{\E}{\operatorname{\mathsf E}}
\newcommand{\PP}{\operatorname{\mathsf P}}$

Without loss of generality (wlog), $c>1$. Let us show a bit more:
\begin{equation}
M_{n,w}:=\inf\{\sum_1^n(x_i y_i)^c\colon (x,y)\in S_{n,w}\}\ge w^{2c-1} \tag{1}
\end{equation}
where
\begin{multline*}
S_{n,w}:=\{(x,y)\colon x=(x_1,\dots,x_n),\ y=(y_1,\dots,y_n),\ x_i\ge0,\ y_i\ge0\ \forall i,\ \\
\sum_1^n x_i\le1,\ \sum_1^n y_i\le1,\ \sum_1^n x_i y_i=w\}.
\end{multline*}
Let $W_n:=\{w\ge0\colon S_{n,w}\ne\emptyset\}$.
Clearly, the inf in (1) is attained for all $w\in W_n$, and that inf equals $\infty$ if $w\notin W_n$ -- in which case (1) is trivial.

Let us prove (1) by induction in $n$. For $n=1$, (1) is obvious. Let $n\ge2$. If $M_n$ is attained at some $x,y$ such that $x_jy_j=0$, then wlog $j=n$ and we may remove the $n$th coordinate from both $x$ and $y$ and thus reduce the problem to showing that $M_{n-1,w}\ge0$. So, by induction, wlog
\begin{equation}
x_iy_i>0\tag{2}
\end{equation}
for all $i=1,\dots,n$. So, wlog the minimizer $(x,y)$ satisfies the Lagrange equations
\begin{equation}
(x_i y_i)^{c-1}y_i=\la y_i+\mu,\quad (x_i y_i)^{c-1}x_i=\la x_i+\nu \tag{3}
\end{equation}
for some real $\la,\mu,\nu$ and all $i$. Multiplying $(x_i y_i)^{c-1}y_i=\la y_i+\mu$ by $x_i$ amd $(x_i y_i)^{c-1}x_i=\la x_i+\nu$ by $y_i$, and then subtracting the results, we get
\begin{equation}
\mu x_i=\nu y_i \tag{4}
\end{equation}
for all $i$.

If $\mu=\nu=0$, then (2) and (3) yield $x_iy_i=w/n$ for all $i$. By Cauchy--Schwarz,
$\sqrt{nw}=\sum_1^n\sqrt{x_iy_i}\le\sqrt{\sum x_i\,\sum y_i}\le1$, so that $nw\le1$. So, $M_{n,w}=\sum_1^n(x_i y_i)^c=w^c/n^{c-1}\ge w^{2c-1}$, as desired.

It remains to consider the case when it is not true that $\mu=\nu=0$. Then, by (4), wlog $y_i=tx_i$ for some $t\in(0,1]$ and all $i$. In this case, inequality (1) follows by H\"older's inequality for the sub-probability measure $\ga$ on the set $\{1,\dots,n\}$, defined by the formula $\ga(\{i\})=x_i$ for all $i$, with $x(i):=x_i$ and $y(i):=y_i=tx_i$:
\begin{multline*}
\sum_1^n x_i y_i=t\int x\,d\ga\le t\Big(\int x^{2c-1}d\ga\Big)^{\frac1{2c-1}}
\Big(\int d\ga\Big)^{\frac{2c-2}{2c-1}} \\
\le t\Big(\int x^{2c-1}d\ga\Big)^{\frac1{2c-1}}
=t\Big(\sum_i x_i^{2c}\Big)^{\frac1{2c-1}} \\
=t^{\frac{c-1}{2c-1}}\Big(\sum_i (x_i y_i)^c\Big)^{\frac1{2c-1}}
\le\Big(\sum_i (x_i y_i)^c\Big)^{\frac1{2c-1}},
\end{multline*}
as desired.

As is oftentimes the case in such situations, it may be possible to distill this proof to get rid of using Lagrange multipliers. On the other hand, using Lagrange multipliers one can basically just compute the answer, with no ingenuity required.