Derivation of the Convexity Relation for Multivariable Function It is easy to derive the convexity relation: if $f(x)$ is convex, $f(t x_{1} + (1 - t) x_{2}) \le t f(x_{1}) + (1 - t) f(x_{2})$ has to be satisfied where $f(x): M \subseteq \mathbb{R} \rightarrow \mathbb{R}$, $x_{1}, x_{2}$ are any two points in $M$ convex set and any $t \in [0, 1]$.
However, I could not figure out how to adapt the same logic to find the right-hand side of the convexity relation for multivariable case if $f(\mathbf{x})$ is convex where $f(\mathbf{x}): C \subseteq \mathbb{R}^{n} \rightarrow \mathbb{R}$, $\mathbf{x}_{1}, \mathbf{x}_{2} \in C$, which $C$ is convex, and $t \in [0, 1]$, it must satisfy the relation below:
$$f(t \mathbf{x}_{1} + (1 - t) \mathbf{x}_{2}) \le t f(\mathbf{x}_{1}) + (1 - t) f(\mathbf{x}_{2}) \tag{1}$$
$t \mathbf{x}_{1} + (1 - t) \mathbf{x}_{2}$ where $t \in [0, 1]$ is an equation of line segment defined within the domain of $f(\mathbf{x})$. Thus, plugging it into the $f(\mathbf{x})$ gives us the corresponding values of $t$ on $f(\mathbf{x})$, on which $\mathbf{x}_{1}$ and $\mathbf{x}_{2}$ are two arbitrary points on $f(\mathbf{x})$. However, I could not get to the right-hand side, $t f(\mathbf{x}_{1}) + (1 - t) f(\mathbf{x}_{2})$. In short, how can I derive the entire relation $(1)$?
Possible(!) Derivation:
Single-variable Case: Let's assume that we have a function $f(x): \mathbb{R} \rightarrow \mathbb{R}$ with a local convexity property within a specific interval, say $I$ and $I \subseteq \mathbb{R}$; that is, if it is a convex function within $I$, any arbitrary two points, say $x_{1}$ and $x_{2}$, we pick from $I$ must satisfy the relation:
$$f(t x_{1} + (1 - t) x_{2}) \le t f(x_{1}) + (1 - t) f(x_{2}) \tag{2}$$
If we draw a line segment with the points within its convexity interval, say $(x_{1}, f(x_{1}))$ and $(x_{2}, f(x_{2}))$ on the function, on $\mathbb{R} \times \mathbb{R}$ plane, this segment will be fully contained in the epigraph of $f(x)$. That is, $f(x) \leq L(x)$ where $L(x)$ is a linear function defined by $(x_{1}, f(x_{1}))$ and $(x_{2}, f(x_{2}))$. If we parametrize points $x$ within between $x_{1}$ and $x_{1}$ by $x = t x_{1} + (1 - t) x_{2}$, we have the set of points defined by $g(t) = f(t x_{1} + (1 - t) x_{2})$, which is nothing but the original $f(x)$ with $t$ parametrization. For the right-hand side of $(2)$ we use the fact that line segment created by $(x_{1}, f(x_{1}))$ and $(x_{1}, f(x_{1}))$ points will be on and above the function $f(x)$. Thus, writing the line equation for $x$,
$$L(x) = \frac{f(x_{1}) - f(x_{2})}{x_{1} - x_{2}} (x - x_{1}) + f(x_{1})$$
If we convert this into $t$ parameterized form by plugging $t x_{1} + (1 - t) x_{2}$ into $x$, we have,
$$
\begin{align*}
l(t) &= \frac{f(x_{1}) - f(x_{2})}{x_{1} - x_{2}} (t x_{1} + (1 - t) x_{2} - x_{1}) + f(x_{1}) \\
&= \frac{f(x_{1}) - f(x_{2})}{x_{1} - x_{2}} ((1 - t) x_{2} - (1 - t)x_{1}) + f(x_{1}) \\
&= \frac{f(x_{1}) - f(x_{2})}{x_{1} - x_{2}} (1 - t) (x_{2} - x_{1}) + f(x_{1}) \\
&= \frac{f(x_{2}) - f(x_{1})}{x_{2} - x_{1}} (1 - t) (x_{2} - x_{1}) + f(x_{1}) \\
&= f(x_{2}) (1 - t) - f(x_{1}) (1 - t) + f(x_{1}) \\
&= f(x_{2}) (1 - t) - f(x_{1}) (1 - t - 1) \\
&= t f(x_{1}) + (1 - t) f(x_{2})
\end{align*}
$$
With $f(x) \leq L(x)$ and $x = t x_{1} + (1 - t) x_{2}$ where $g(t) = f(t x_{1} + (1 - t) x_{2})$ $l(t) = t f(x_{1}) + (1 - t) f(x_{2})$ and we have,
$$g(t) \leq l(t) \implies f(t x_{1} + (1 - t) x_{2}) \leq t f(x_{1}) + (1 - t) f(x_{2})\ \blacksquare$$
Multivariable Case: Let $f(\mathbf{x}): \mathbb{R}^{n} \rightarrow \mathbb{R}$ and $x_{1}, x_{2} \in \mathbb{R}^{n}$. Now let's define $F(t)$ as,
$$f(t \mathbf{x_{1}} + (1 - t) \mathbf{x_{2}}) = F(t)$$
That is, $f(\mathbf{x})$ is parametrized as another function $F(t)$, which is single-variable. Then applying $(2)$ for $F(t)$ would mean deriving the convexity condition $(2)$ for $F(t)$ between $t = [0,1]$ where,
$$t = 0 \implies f(\mathbf{x_{2}}) = F(0)$$
$$ t = 1 \implies f(\mathbf{x_{1}}) = F(1)$$
Replacing $x_{1}$ with $1$ and $x_{2}$ with $0$ in $(2)$ above and if we parametrize $t = [0, 1]$ with $\lambda$ as in $t = 0 + \lambda(1 - 0) \implies (t = \lambda)$, we have,
$$
\begin{align*}
F(\lambda 1 + (1 - \lambda)0) &\leq \frac{F(1) - F(0)}{1 - 0} (\lambda - 0) + F(0) \\
F(\lambda) &\leq (F(1) - F(0)) (\lambda - 0) + F(0) \\
F(t) &\leq F(1) t + (1 - t) F(0) \\
\end{align*}
$$
With $f(t \mathbf{x_{1}} + (1 - t) \mathbf{x_{2}}) = F(t)$, $f(\mathbf{x_{2}}) = F(0)$ and $f(\mathbf{x_{1}}) = F(1)$ we have,
$$f(t \mathbf{x_{1}} + (1 - t) \mathbf{x_{2}}) \leq t f(\mathbf{x_{1}}) + (1 - t) f(\mathbf{x_{2}})\ \blacksquare$$
EDIT: (!) was appended.
 A: Based on the comments, it looks like the claim is
\begin{equation}
\text{If}\;\;f\colon\mathbb{R}^N\to\mathbb{R}\;\;\text{is convex in every component, then}\;\;f\;\;\text{is convex on}\;\;\mathbb{R}^N,
\end{equation}
which is not true.
Counterexample: let $f\colon\mathbb{R}^2\to\mathbb{R}\colon (\xi_1,\xi_2)\mapsto\xi_1\xi_2$. In each component (i.e. when one component is a free variable while the other component is held constant), $f$ is actually linear, hence it is convex. Set ${\bf x_1}=(-1,1)$, ${\bf x_2}=(1,-1)$, and $\alpha=1/2$. Then
$$0=f(\alpha{\bf x_1} + (1-\alpha){\bf x_2}) > \alpha f({\bf x_1})+(1-\alpha)f({\bf x_2})=-1,$$
so when both variables are free, we find $f$ is nonconvex.
EDIT: Apparently the claim was supposed to be
\begin{equation}
\text{The epigraph of}\;\;f\colon\mathbb{R}^N\to\mathbb{R}\;\;\text{is convex}\;\;\Leftrightarrow\;\;\text{(1) holds,}
\end{equation}
which is a standard exercise in convex analysis. You can find this proof in, e.g. Bauschke & Combettes' book (vol. 2), Proposition 8.4.
