I do not think the problem is very difficult : Wlog, we may assume that [ABC] = 1. Let A', B' and C' be the midpoints of the sides BC, CA and AB respectively. If R is on the segment AC' and Q is on the segment AB', then clearly [AQR] \leq [AB'C'] = 1/4, so we are done in that case. We have similar conclusion in the case where two of the points P,Q,R lies on the half-sides from the common vertex. Thus, we may assume that R is on the segment BC', Q is on AB' and P is on CA' (the case R on AC', Q on CB' and P on BA' is similar). Let's consider R and P to be fixed. From ouir choices of the points, the line parallel to PR through B' meets the side AB in a point which belongs to the segment AC' (look about the respective slope). Thus, for any point in the segment AB', the distance from this point to the line PR is at least the distance from B' to the line PR. It follows that [PQR] \geq [B'PR]. With a similar reasoning, we prove that : [PQR] \geq [B'PR] \geq [B'C'P] \geq [A'B'C'] = 1/4. Since [AQR]+[BRP]+[CPQ]+[RPQ] = [ABC] = 1, we deduce that [AQR]+[BRP]+[CRQ] \leq 3/4, from which the result follows easily. Pierre.

Pierre gave too complicated solution Let $P$, $Q$, $R$ divide sides in ratios $a:1-a$, $b:1-b$ and $c:1-c$ respectively, then if we suppose contrary it will mean that $a(1-b)>1/4$, $b(1-c)>1/4$, $c(1-a)>1/4$. Multiplying them we obtain $abc(1-a)(1-b)(1-c)>1/64$, though $x(1-x)\leq 1/4$ for all $x\in[0,1]$. Contradiction.

Myth wrote: Pierre gave too complicated solution 'Too complicated' is my third name. ('Too long' is my second) [Moderator edit: Another solution to this problem is to conclude it from the stronger problem at http://www.mathlinks.ro/Forum/viewtopic.php?t=5179 .] Pierre.

We note that the equality is attained if P,Q,R are the middle points of the sides. Now we define $\overline{AR}=a$, $\overline{BR}=b$, $\overline{BP}=c$, $\overline{CP}=d$, $\overline{CQ}=e$, $\overline{AQ}=f$. Now we suppose that $(\triangle ARQ)>\frac{(\triangle ABC)}{4}$, $(\triangle BRP)>\frac{(\triangle ABC)}{4}$, $(\triangle CPQ)>\frac{(\triangle ABC)}{4}$ Now we have $\frac{af\cdot sen\angle BAC}{2}>\frac{\frac{(a+b)(e+f)\cdot sen\angle BAC}{2}}{4}$ $\frac{bc\cdot sen\angle ABC}{2}>\frac{\frac{(a+b)(c+d)\cdot sen\angle ABC}{2}}{4}$ $\frac{de\cdot sen\angle BCA}{2}>\frac{\frac{(c+d)(e+f)\cdot sen\angle BCA}{2}}{4}$ Simplifying and adding the inequalities we have $abcdef>\frac{(a+b)^2(c+d)^2(e+f)^2}{64}$ that is a contradiction, because for AM-GM over abd, afd, bcd, cdf, abe, aef, cbe, cfe we have $\frac{abd+afd+bcd+cdf+abe+aef+cbe+cfe}{8}\ge \sqrt[8]{a^4b^4c^4d^4e^4f^4}$ $\frac{(a+b)^2(c+d)^2(e+f)^2}{64}\ge abcdef$ we finished )