2nd part: take $f(x)=|x|$

Let $P(x,y)$ denotes the assertion that $|f(x+y)|\ge |f(x)+f(y)|$. Note first that $P(0,0)$ yields $f(0)=0$. Further, $P(x,-x)$ yields $0\ge |f(x)+f(-x)|$, thus $f(-x)=-f(x)$ for all $x$. Now, we inspect $P(x+y,-x)$ and $P(x+y,-y)$ to get \begin{align*} |f(x)|\ge |f(x+y)-f(x)| \implies f(x)^2 &\ge f(x+y)^2 -2f(x+y)f(y)+f(y)^2 \\ |f(y)|\ge |f(x+y)-f(x)| \implies f(y)^2 &\ge f(x+y)^2 -2f(x+y)f(x)+f(x)^2. \end{align*}Summing these up, we find that \[ f(x+y)\bigl(f(x)+f(y)\bigr)\ge f(x+y)^2. \]In particular, for every $x,y$, ${\rm sgn}(f(x+y)) = {\rm sgn}(f(x)+f(y))$. We are now ready for conclusion. Assume first $f(x+y)\ge 0$. Then, $f(x)+f(y)\ge f(x+y)\ge 0$. So, $|f(x)+f(y)| = f(x)+f(y) \ge f(x+y)=|f(x+y)|\ge |f(x)+f(y)|$, yielding the desired equality. Likewise, if $f(x+y)\le 0$, then we find $f(x)+f(y)\le f(x+y) \le 0$, forcing $|f(x)+f(y)|\ge |f(x+y)|$, yielding the conclusion.