if the sum of the digits of $a+b$, $b+c$ and $a+c$ is $<5$, then is the sum of the digits of $a+b+c$ less or equal than 50? This is a problem from a math olympiad: 
Let $S(x)$ denote the sum of the digits of a natural number $x$ in its decimal representation. Do there exist three natural numbers $a$, $b$ and $c$ with $S(a+b)<5$, $S(a+c)<5$, $S(b+c)<5$ and $S(a+b+c)>50$?
My attempt: 
NOTATION: Given a real number $x$, $[x]$ is the integer part of $x$.
REMARK 1: Let $x$ be a natural number with $2$ digits. Then the sum of its two digits is 
$$ \left[\frac{x}{10}\right]+x-\left[\frac{x}{10}\right]\cdot 10=x-9\cdot \left[\frac{x}{10}\right]. $$
REMARK 2: If $x$ is a real number with $\left[ x\right]\geq1$, then $\left[ 2x\right]\leq 3\left[ x\right]$.
Let $a$, $b$ and $c$ be any natural numbers. Write
$$ a=\sum_{i=0}^{n-1}a_i 10^i,\quad b=\sum_{i=0}^{n-1}b_i 10^i, \quad c=\sum_{i=0}^{n-1}c_i 10^i, $$
where $a_i,b_i,c_i\in\{0,\ldots,9\}$ and $a_n\neq0$ or $b_n\neq0$ or $c_n\neq0$. By Remark 1,
$$ S(a+b+c)=\sum_{i=0}^{n-1}\left(a_i+b_i+c_i-9\cdot \left[ \frac{a_i+b_i+c_i}{10}\right]\right), $$
$$ S(a+b)=\sum_{i=0}^{n-1}\left(a_i+b_i-9\cdot \left[ \frac{a_i+b_i}{10}\right]\right), $$
$$ S(b+c)=\sum_{i=0}^{n-1}\left(b_i+c_i-9\cdot \left[ \frac{b_i+c_i}{10}\right]\right), $$
$$ S(a+c)=\sum_{i=0}^{n-1}\left(a_i+c_i-9\cdot \left[ \frac{a_i+c_i}{10}\right]\right). $$
By Remark 2, if $\left[ (a_i+b_i+c_i)/10\right]\geq1$, then $$3\left[(a_i+b_i+c_i)/10\right]\geq \left[ 2(a_i+b_i+c_i)/10\right]\geq \left[ (a_i+b_i)/10\right]+\left[ (b_i+c_i)/10\right]+\left[ (a_i+c_i)/10\right].$$ Thus,
$$ \begin{align*}
S(a+b+c)= {} & \sum_{i=0}^{n-1}\left(a_i+b_i+c_i-9\cdot \left[ \frac{a_i+b_i+c_i}{10}\right]\right) \\\\
= {} & \frac12\sum_{i=0}^{n-1}(a_i+b_i)+\frac12\sum_{i=0}^{n-1}(b_i+c_i)+\frac12\sum_{i=0}^{n-1}(a_i+c_i)\\
&\qquad-9\sum_{\substack{{0\leq i\leq n-1}\\{(a_i+b_i+c_i)/10\geq1}}}\left[ \frac{a_i+b_i+c_i}{10}\right] \\\\
\leq {} & \frac12\sum_{i=0}^{n-1}(a_i+b_i)+\frac12\sum_{i=0}^{n-1}(b_i+c_i)+\frac12\sum_{i=0}^{n-1}(a_i+c_i)-3\sum_{i=0}^{n-1}\left[ \frac{a_i+b_i}{10}\right]\\
&\qquad-3\sum_{i=0}^{n-1}\left[ \frac{b_i+c_i}{10}\right]-3\sum_{i=0}^{n-1}\left[ \frac{a_i+c_i}{10}\right] \\\\
= & \frac12\sum_{i=0}^{n-1}\left(a_i+b_i-6\left[ \frac{a_i+b_i}{10}\right]\right)+\frac12\sum_{i=0}^{n-1}\left(b_i+c_i-6\left[ \frac{b_i+c_i}{10}\right]\right)\\
&\qquad+\frac12\sum_{i=0}^{n-1}\left(a_i+c_i-6\left[ \frac{a_i+c_i}{10}\right]\right) \\\\
= {} & \frac{S(a+b)+S(b+c)+S(a+c)}{2}+\frac{3}{2}\sum_{i=0}^{n-1}\left[ \frac{a_i+b_i}{10}\right]\\
&\qquad+\frac{3}{2}\sum_{i=0}^{n-1}\left[ \frac{b_i+c_i}{10}\right]+\frac{3}{2}\sum_{i=0}^{n-1}\left[ \frac{a_i+c_i}{10}\right] \\\\
\leq {} & \frac{S(a+b)+S(b+c)+S(a+c)}{2}+\frac{3}{2}S(a+b)+\frac{3}{2}S(b+c)+\frac{3}{2}S(a+c) \\\\
= {} & 2(S(a+b)+S(b+c)+S(a+c))\;.
\end{align*}
$$ Then, if $S(a+b)<5$, $S(a+c)<5$, $S(b+c)<5$, we have $S(a+b+c)<2(5+5+5)=30$. 

The answer to the question in the title is no -- it is possible that $S(a+b+c) = 51$. In particular, your proof is incorrect.
Take
\begin{align*}
a &= 545\;545\;545\;5 \\
b &= 554\;554\;554\;5 \\
c &= 455\;455\;455\;5.
\end{align*}
(Spaces to aid readability, so you can tell what's going on here.) Then we have
\begin{align*}
a+b &= 1\;100\;100\;100\;0 \\
a+c &= 1\;001\;001\;001\;0 \\
b+c &= 1\;010\;010\;010\;0 \\
\end{align*}
So $S(a+b) = S(a+c) = S(b+c) = 4 < 5$.
But
$$
a+b+c = 1\;555\;555\;555\;5
$$
so $S(a+b+c) = 51$.
Why your proof fails
The error is in the beginning, in asserting that the sum of digits of $S(a+b+c)$ is equal to $\sum_i S(a_i + b_i + c_i)$, and similar for $a+b$,$a+c$, and $b+c$. While $S(a+b+c) \le \sum_i S(a_i + b_i + c_i)$, they certainly need not be equal. Replacing it with $\le$ instead of $=$, you can see that your proof no longer goes through.
Also note that if you use $S(a+b), S(a+c), S(b+c) \le 4$ at the end instead of $S(a+b), S(a+c), S(b+c) < 5$, then your proof would purport to show $S(a+b+c) \le 2 (12) = 24$, which has even simpler counterexamples.
Additional notes
While it is not true that $S(a+b+c) \le 50$, it is still true that $S(a+b+c) \le 60$. Here is a proof.
First, we can note that
$$
S(x + y) \le S(x) + S(y). \tag{1}
$$
To prove it, first imagine adding $x$ and $y$ in base 10 without carrying. Then the sum of digits stays the same. But each time we carry, we reduce the sum. So $x+y$ after all carrying is completed must have a smaller sum of digits.
A second observation is that
$$
S(2x) \ge \frac15 S(x) \tag{2}
$$
To prove it, first consider what multliplication by $2$ does to single digits: $1 \mapsto 2$, $2 \mapsto 4$, $3 \mapsto 6$, $4 \mapsto 8$, $5 \mapsto 10$, $6 \mapsto 12$, $7 \mapsto 14$, $8 \mapsto 16$, $9 \mapsto 18$. 
In every case, the sum of digits is at least $\frac{1}{5}$ of the original digit (the worst case being the digit $5$ becoming $10$).
Then notice (and this is special to multiplication by $2$ in particular) that to find $2x$, we may multiply each digit by $2$ individually, resulting in a one or two digit number for each digit, and then add all these up with no carrying.
In other words, $S(2x)$ is the sum over digits $d$ of $x$ of $S(2d)$. Since $S(2d) \ge \frac{1}{5}$, it follows that $S(2x) \ge \frac{1}{5} S(x)$.
Putting (2) and (1) together, in that order:
$$
S(a+b+c) \le 5S(2(a+b+c)) \le 5\left[S(a+b) + S(b+c) + S(a+c)\right] \le 5\left[4 + 4 + 4\right] = 60.
$$
