EDIT: Part 4 added. EDIT2: Second proof of Part 4 added.
Theorem 1. If $V_1$, $V_2$, ..., $V_m$ are $m$ pairwise nonisomorphic irreducible representations of a finite-dimensional algebra $A$ over a field $k$ (not necessarily algebraically closed, not necessarily of characteristic $0$), then $\dim V_1+\dim V_2+...+\dim V_m\leq\dim A$.
ProofFirst proof of Theorem 1. At first, for every $i\in\left\lbrace 1,2,...,m\right\rbrace$, the (left) representation $V_i^{\ast}$ of the algebra $A^{\mathrm{op}}$ (this representation is defined by $a\cdot f=\left(v\mapsto f\left(av\right)\right)$ for any $f\in V_i^{\ast}$ and $a\in A$) is irreducible (since $V_i$ is irreducible) and therefore isomorphic to a quotient of the regular (left) representation $A^{\mathrm{op}}$ (since we can choose some nonzero $u\in V_i^{\ast}$, and then the map $A^{\mathrm{op}}\to V_i^{\ast}$ given by $a\mapsto au$ must be surjective, because its image is a nonzero subrepresentation of $V_i^{\ast}$ and therefore equal to $V_i^{\ast}$ due to the irreducibility of $V_i^{\ast}$). Hence, by duality, $V_i$ is isomorphic to a subrepresentation of the (left) representation $A^{\mathrm{op}\ast}=A^{\ast}$ of $A$. Hence, from now on, let's assume that $V_i$ actually is a subrepresentation of $A^{\ast}$ for every $i\in\left\lbrace 1,2,...,m\right\rbrace$.
Now, according to Theorem 2.2 and Remark 2.3 of Etingof's "Introduction to representation theory", any subrepresentation of the direct sum $V_1\oplus V_2\oplus ...\oplus V_{m-1}$ must be a direct sum of the form $r_1V_1\oplus r_2V_2\oplus ...\oplus r_{m-1}V_{m-1}$ for some nonnegative integers $r_1$, $r_2$, ..., $r_{m-1}$. Hence, every irreducible subrepresentation of the direct sum $V_1\oplus V_2\oplus ...\oplus V_{m-1}$ must be one of the representations $V_1$, $V_2$, ..., $V_{m-1}$. Since we know that $V_m$ is isomorphic to a subrepresentation of the direct sum $V_1\oplus V_2\oplus ...\oplus V_{m-1}$, we conclude that $V_m$ is isomorphic to one of the representations $V_1$, $V_2$, ..., $V_{m-1}$. This contradicts the non-isomorphy of the representations $V_1$, $V_2$, ..., $V_m$. Thus, we have proven that the sum $V_1+V_2+...+V_m$ is actually a direct sum. Consequently, $\dim V_1+\dim V_2+...+\dim V_m=\dim\left(V_1+V_2+...+V_m\right)\leq \dim A^{\ast}=\dim A$, qedand Theorem 1 is proven.
Second proof of Theorem 1. I just learnt the following simpler proof of Theorem 1 from §1 Lemma 1 in Crawley-Boevey's "Lectures on representation theory and invariant theory":
Let $0=A_0\subseteq A_1\subseteq A_2\subseteq ...\subseteq A_k=A$ be a composition series of the regular representation $A$ of $A$. Then, by the definition of a composition series, for every $i\in \left\lbrace 1,2,...,k\right\rbrace$, the representation $A_i/A_{i-1}$ of $A$ is irreducible.
Let $T$ be an irreducible representation of $A$. We are going to prove that there exists some $I\in \left\lbrace 1,2,...,k\right\rbrace$ such that $T\cong A_I/A_{I-1}$ (as representations of $A$).
In fact, let $I$ be the smallest element $i\in \left\lbrace 1,2,...,k\right\rbrace$ satisfying $A_iT\neq 0$ (such elements $i$ exist, because $A_kT=AT=T\neq 0$). Then, $A_IT\neq 0$, but $A_{I-1}T=0$. Now, choose some vector $t\in T$ such that $A_It\neq 0$ (such a vector $t$ exists, because $A_IT\neq 0$), and consider the map $f:A_I\to T$ defined by $f\left(a\right)=at$ for every $a\in A_I$. Then, this map $f$ is a homomorphism of representations of $A$. Since it maps the subrepresentation $A_{I-1}$ to $0$ (because $f\left(A_{I-1}\right)=A_{I-1}t\subseteq A_{I-1}T=0$), it gives rise to a map $g:A_I/A_{I-1}\to T$, which, of course, must also be a homomorphism of representations of $A$. Since $A_I/A_{I-1}$ and $T$ are irreducible representations of $A$, it follows from Schur's lemma that any homomorphism of representations from $A_I/A_{I-1}$ to $T$ is either an isomorphism or identically zero. Hence, $g$ is either an isomorphism or identically zero. But $g$ is not identically zero (since $g\left(A_I/A_{I-1}\right)=f\left(A_I\right)=A_It\neq 0$), so that $g$ must be an isomorphism, i. e., we have $T\cong A_I/A_{I-1}$.
So we have just proven that
(1) For every irreducible representation $T$ of $A$, there exists some $I\in \left\lbrace 1,2,...,k\right\rbrace$ such that $T\cong A_I/A_{I-1}$ (as representations of $A$).
Denote this $I$ by $I_T$ in order to make it clear that it depends on $T$. So we have $T\cong A_{I_T}/A_{I_T-1}$ for each irreducible representation $T$ of $A$. Applying this to $T=V_i$ for every $i\in\left\lbrace 1,2,...,m\right\rbrace$, we see that $V_i\cong A_{I_{V_i}}/A_{I_{V_i}-1}$ for every $i\in\left\lbrace 1,2,...,m\right\rbrace$. Hence, the elements $I_{V_1}$, $I_{V_2}$, ..., $I_{V_m}$ of the set $\left\lbrace 1,2,...,k\right\rbrace$ are pairwise distinct (because $I_{V_i}=I_{V_j}$ would yield $V_i\cong A_{I_{V_i}}/A_{I_{V_i}-1}=A_{I_{V_j}}/A_{I_{V_j}-1}\cong V_j$, but the representations $V_1$, $V_2$, ..., $V_m$ are pairwise nonisomorphic), and thus
$\sum\limits_{i=1}^{m}\dim\left(A_{I_{V_i}}/A_{I_{V_i}-1}\right)=\sum\limits_{\substack{j\in\left\lbrace 1,2,...,k\right\rbrace ;\ \\ \text{there exists }\\ i\in\left\lbrace 1,2,...,m\right\rbrace \\ \text{ such that }j=I_{V_i}}}\dim\left(A_j/A_{j-1}\right)$
$\leq \sum\limits_{j\in\left\lbrace 1,2,...,k\right\rbrace}\dim\left(A_j/A_{j-1}\right)$ (since $\dim\left(A_j/A_{j-1}\right)\geq 0$ for every $j$, so that adding more summands cannot decrease the sum)
$=\sum\limits_{j=1}^{k}\dim\left(A_j/A_{j-1}\right)=\sum\limits_{j=1}^{k}\left(\dim A_j-\dim A_{j-1}\right)$.
Since $\dim\left(A_{I_{V_i}}/A_{I_{V_i}-1}\right)=\dim V_i$ for each $i$ (due to $A_{I_{V_i}}/A_{I_{V_i}-1}\cong V_i$) and $\sum\limits_{j=1}^{k}\left(\dim A_j-\dim A_{j-1}\right)=\dim A$ (in fact, the sum $\sum\limits_{j=1}^{k}\left(\dim A_j-\dim A_{j-1}\right)$ is a telescopic sum and simplifies to $\dim A_k-\dim A_0=\dim A-\dim 0=\dim A-0=\dim A$), this inequality becomes $\sum\limits_{i=1}^{m}\dim V_i\leq\dim A$. This proves Theorem 1.
