We solve the problem in two steps: first, we show that of the two points on one side, one of them can be moved into a vertex such that the area of all four triangles increases, and we finish the problem by showing that the area of all the four resulting triangles cannot be each greater than $1/4$. It will be convenient to use different notation for the two parts. We will also show that equality holds (i.e. the minimal triangle has area $1/4$) exactly when the four points selected are: a vertex $Z$, two midpoints, and a point on segment $\overline{XY}$, where $X$ is the third midpoint, of side $ZY$. For the first part, without loss of generality, we have the following configuration: there are two points on side $BC$, namely $D,E$ with $B,D,E,C$ in this order, and the other two points are $P\in CA$ and $Q\in AB$ such that $PQ$ intersects line $BC$ on the ray $CB$ at $X$. This way, if we move $E$ to $C$ on $\overline{DC}$, $d(E,PQ)$ increases because $\frac{d(E,PQ)}{d(C,PQ)}=\frac{EX}{CX}\le 1$. Moreover, $d(E,PD)$ and $d(E,QD)$ increase because $\frac{d(E,PD)}{d(C,PD)}=\frac{ED}{CD}=\frac{d(E,QD)}{d(C,QD)}$ due to similarity. Hence if we move $E$ to $C$, the area of $\triangle EPQ$, $\triangle EPD$, $\triangle EQD$ may not decrease, and the fourth triangle $DPQ$ remains the same. $\blacksquare$ (Note that if $E,\neq C$, then there cannot be equality.) So now without loss of generality, among the $P_i$ we have one vertex and one on each side of the triangle. Draw a new diagram, let the selected vertex be $A$, let the points on side $AB,AC$ be $X,Y$ and let the points on $BC$ be $P$. Introduce the following notations: $\frac{AX}{AB}=x$, $\frac{AY}{AC}=y$, $\frac{BP}{BC}=\alpha$, $\frac{CP}{CB}=\beta$. Then we can easily find that $A_1=[AXY]=xy$, $A_2=[XYP]=1-xy-\alpha(1-x)-\beta(1-y)$, $A_3=[APX]=\alpha x$, $A_4=[BPY]=\beta y$. Note that $x,y,\alpha,\beta\in[0;1]$ and $\alpha+\beta=1$. Suppose all the $A_i$ are $\ge 1/4$ (we'll see that all four cannot be $>1/4$, and what we deduce will be exactly the case of equality). Studying the case of equality, we are motivated to consider the following two inequalities: $$1\le 4A_2=4-4xy-\underbrace{4(\alpha+\beta)}_{=4}+4(\alpha x+\beta y),$$ $$0\le (4A_3-1)(4A_4-1)=16\alpha\beta xy+1-4(\alpha x+\beta y).$$ Adding the two, we arrive at $$1\le -4xy+16\alpha\beta xy+1,$$ hence $4xy\le 16\alpha\beta xy$, which we can divide by $4xy$ (which is positive as $A_1\ge 1/4$) to get $1\le 4\alpha\beta$. But also $4\alpha\beta\le (\alpha+\beta)^2=1$. This yields a contradiction if even $A_2>1/4$. $\blacksquare$ But if we allow $=1/4$, then for equality to hold everywhere, we need $\alpha=\beta=1/2$ (so $P$ must be a midpoint), and for $A_2$ and one of $A_3,A_4$ to be $=1/2$. Note that $A_1=xy\ge 1/4$ implies that $x$ or $y$ is $\ge 1/2$; say if $x\ge 1/2$, then $A_3=1/4$ is only possible with $x=1/2$, which leaves us with $y\ge 1/2$ due to $xy\ge 1/4$. It is easy to check that these indeed give equality.