67th Romanian Mathematical Olympiad

a) Let $t=\frac{d-b}{a-d}=\frac{c-a}{a-d}\in [0,1]$, then $b=ta+(1-t)d$ and $c=(1-t)a+td$. Using (P2) yields $$f(b)=f(ta+(1-t)d)\le tf(a)+(1-t)f(d).$$ Similarly, $f(c)\le (1-t)f(a)+tf(d)$. Adding up these two inequalities gives the result.

b) We will prove the inequality by induction and we start with the base case $n=3$. Let $x,y,z$ be arbitrary real numbers. Two of them, say $x,y$, have the same sign. If $x\ge 0$ then $z+(x+y+z)=(x+z)+(y+z)$ and $z\le x+z,y+z\le x+y+z$. We deduce from a) that $$f(x+z)+f(y+z)\le f(z)+f(x+y+z).$$ Adding this with $f(x+y)\le f(x)+f(y)$ yields the result.

If $x<0$, then $x+y+z\le y+z$, $z+x\le z$ and we proceed in a similar manner. Now, assume the assertion true for $n$ and consider the numbers $x_1,x_2,\dots ,x_{n+1}$. Applying the inductive hypothesis to the numbers $x_1,x_2,\dots ,x_{n-1}$ and $x_n+x_{n+1}$ leads to $$f(x_1+\cdots +x_n+x_{n+1})+(n-2)(f(x_1)+\cdots +f(x_{n-1})+f(x_n+x_{n+1}))\ge \sum _{1\le i<j\le n-1}f(x_i+x_j)+\sum _{1\le i\le n-1}f(x_i+x_n+x_{n+1}).$$ The conclusion is obtained by adding up the inequalities $f(x_i+x_n+x_{n+1})\ge f(x_i+x_n)+f(x_i+x_{n+1})+f(x_n+x_{n+1})-f(x_i)-f(x_n)-(x_{n+1})$.