At first we look at the geometrical situation of the convex hull of three points in general position in $\mathbb{R^2}$.
Convex hull of three points with $(a,b)$ inside:
In OPs referenced article we have a triangle with points $(u_j,v_j),1\leq j\leq 3$ and a point $(a,b)$ inside the triangle as shown in the graphic below:
$\qquad\qquad\quad$
We can write $(a,b)$ as convex linear combination of $(u_j,v_j)$ according the graphic as follows:
\begin{align*}
\binom{a^{\prime}}{b^{\prime}}&=\binom{u_2}{v_2}+\lambda\left[\binom{u_3}{v_3}-\binom{u_2}{v_2}\right]\\
&=(1-\lambda)\binom{u_2}{v_2}+\lambda\binom{u_3}{v_3}\\
\\
\color{blue}{\binom{a}{b}}&=\binom{u_1}{v_1}+\mu\left[\binom{a^{\prime}}{b^{\prime}}-\binom{u_1}{v_1}\right]\\
&=(1-\mu)\binom{u_1}{v_1}+\mu\binom{a^{\prime}}{b^{\prime}}\\
&=(1-\mu)\binom{u_1}{v_1}+\mu\left[(1-\lambda)\binom{u_2}{v_2}+\lambda\binom{u_3}{v_3}\right]\\
&\,\,\color{blue}{=(1-\mu)\binom{u_1}{v_1}+\mu(1-\lambda)\binom{u_2}{v_2}+\lambda\mu\binom{u_3}{v_3}}\tag{1}
\end{align*}
Note we have in (1) a convex combination of the points $(u_j,v_j)$ since
\begin{align*}
(1-\mu)+\mu(1-\lambda)+\lambda\mu=1
\end{align*}
Convex hull of three points with $(a,b)=(1,1)$ at the boundary:
Now we consider the triangle with the three points $(0,0), (p,0), (0,q)$, $p>1, q>1$ and the point $(a,b)=(1,1)$ at the line segment from $(p,0)$ to $(0,q)$.
$\qquad\qquad\qquad$
We calculate similarly as above:
\begin{align*}
\color{blue}{\binom{1}{1}}&=\binom{p}{0}+\nu\left[\binom{0}{q}-\binom{p}{0}\right]\\
&\,\,\color{blue}{=(1-\nu)\binom{p}{0}+\nu\binom{0}{q}}
\end{align*}
We observe in this special case where $(a,b)=(1,1)$ is at the boundary of the convex hull, we need only two points $(p,0)$ and $(0,q)$ to obtain a convex combination of the triangle for $(1,1)$.
Hölder's inequality:
A few words to Hölder's inequality which might be helpful. We have for positive real numbers $x$ and $y$ and $p>1, q>1$ real numbers with $\frac{1}{p}+\frac{1}{q}=1$ the arithmetic-geometric means inequality:
\begin{align*}
x^{\frac{1}{p}}y^{\frac{1}{q}}\leq \frac{1}{p}x+\frac{1}{q}y\tag{2}
\end{align*}
We define for a $p$-integrable measurable function $f$:
\begin{align*}
N_p(f):=\left(\int|f|^p\,d\mu\right)^{\frac{1}{p}}
\end{align*}
and take
\begin{align*}
x:=\left(\frac{|f(\omega)|}{N_p(f)}\right)^p,\quad y:=\left(\frac{|g(\omega)|}{N_q(g)}\right)^q\tag{3}
\end{align*}
We obtain from (2) and (3):
\begin{align*}
\frac{|f\,g|}{N_p(f)N_q(g)}&\leq \frac{1}{p\left(N_p(f)\right)^p}|f|^p+\frac{1}{q\left(N_q(g)\right)^q}|g|^q\tag{4}\\
\frac{1}{N_p(f)N_q(g)}\int|f\,g|\,d\mu
&\leq \frac{1}{p\left(N_p(f)\right)^p}\int|f|^p\,d\mu+\frac{1}{q\left(N_q(g)\right)^q}\int|g|^q\,d\mu\\
&=\frac{1}{p}+\frac{1}{q}\\
&=1\\
\color{blue}{N_1(fg)}&\color{blue}{\leq N_p(f)N_q(g) }\tag{5}
\end{align*}
and (5) gives Hölder's inequality.
Note the transformation from the arithmetic mean in (4) to the multiplication of $N_p(f)$ with $N_q(g)$ in (5).
