Let $P(a,b,c,d)$ denote the given assertion. $P(0,0,0,0) \implies f(0)=0$. $P(a,a,c,c) \implies \color{blue}{f(a-c)^2 \le (a-c)f(a-c)}$ so if $a-c < 0 \implies f(a-c) \le 0$ and $f(a-c) > 0 \implies f(a-c) \ge 0$ otherwise $f(a-c)^2 < 0$. $P(a,b,c,b) \implies f(a-b) [f(c-b) + f(b-c) ] \le 0 \implies f(y)(f(x) + f(-x)) \le 0$. If $f(x) + f(-x) \neq 0$ then we can select suitable $y>0$ or $y <0$ to obtain a contradiction so $f(-x) = -f(x)$. $P(a,b,b,a) \implies f(a-b)f(b-a) \le (a-b)f(b-a) \implies -f(a-b)^2 \le -(a-b)f(a-b) \implies \color{red}{f(a-b)^2 \ge (a-b)f(a-b)}$. Combining the coloured inequalities we have $f(x) \in \{0,x\}$. For pointwise trap, assume $f(a)=0$ and $f(b) =b$ for $a,b \neq 0$. $P(a,b,0,0) \implies 0\le ab$ and $P(a,0,0,b) \implies 0 \le -ab$ so $ab=0$ which is a contradiction, yielding the solutions $f(x)=0$ for all $x$ or $f(x)=x$ for all $x$.