BxMO Team Selection Test

The solutions to the given functional inequality are $f(x)=0$ for all $x$ and $f(x)=x$ for all $x$. For $f(x)=0$, we easily find that equality always holds. For $f(x)=x$ we check that $$\begin{array}{rl}(a-b)(c-d)+(a-d)(b-c)& =ac-ad-bc+bd+ab-ac-bd+cd\\ & =-ad-bc+ab+cd\\ & =(a-c)(b-d),\end{array}$$ so equality holds in that case too. Now we show that these functions are the only two solutions.

Substituting $a=b=c=d=0$ gives us $2f(0{)}^2\le 0$, and therefore $f(0)=0$. Then we substitute $b=a-x$, $c=a$, and $d=a-y$, which gives $a-b=x$, $a-c=0$ and $a-d=y$. From this we deduce that $$f(y)(f(x)+f(-x))\le 0(1)$$ Suppose there is a $y$ such that $f(y)\ne 0$. If we then substitute $x=y$ in the equation above and move one of the terms to the right, we find that $$0<f(y{)}^2\le -f(y)f(-y).$$ Therefore one of the two values $f(y)$ and $f(-y)$ is positive and the other is negative. Assume without loss of generality that $f(y)$ is positive. Now, given arbitrary $a$ and $y$, substitute $b=a$, $c=0$, and $d=a-y$. Then we find that $f(y)f(a)\le af(y)$. Thus, if we divide both sides by the positive $f(y)$, we get $f(a)\le a$. On the other hand, if we substitute $b=a$, $c=0$, and $d=a+y$, we find that $f(-y)f(a)\le af(-y)$. Since $f(-y)$ is negative, the sign flips when we divide by $f(-y)$ so we deduce that $f(a)\ge a$. We conclude that $f(a)=a$ for all real $a$. Therefore a solution $f$ of the given functional inequality is either the zero function or $f(a)=a$ for all real $a$, and we confirmed in the beginning that both are indeed solutions. $\square$