For simplifying the notations, denote $y$ the longest altitude of $\triangle{MNP}$. Also denote $H$ the orthocenter of $\triangle{MNP}$ and $\triangle{A'B'C'}$ the pedal triangle of $H$. Thus $\frac{\sum{HA'}}{x}\geq \sum{\frac{HA'}{h_{a}}}= \sum{\frac{[HBC]}{[ABC]}=1}$ Therefore $x\leq \sum{HA'}$ $(1)$. Now let's see that $\sum{HA'}\leq \sum{HM}$, and try to prove that $\sum{HM}\leq 2y$ $(2)$. Suppose that $MM_{1}$ is the longest altitude of $\triangle{MNP}$, therefore $(NP)$ is the shortest side. Now call $D$ the symmetrical point of $H$, w.r.t to $M_{1}$. Thus $D$ is on the circumcircle of $\triangle{MNP}$. Observe that $\sum{HM}= HM+DN+DP$, and rewrite $(2)$ as: $DN+DP \leq MD$, which is obvious by the Ptolemy theorem in the cyclic quadrilateral $MNDP$ ($MD \cdot NP = DN \cdot MP+DP \cdot MN$, but $NP = \min\left\{NP,PM,MN\right\}$), so $(2)$ is proved. Thus by $(1)$ and $(2)$ we have that $x\leq 2y$. Remark: Another nice solution involves an homothety and a translation, I will post it after a month.

An approach without using orthocenters...

Scrambled wrote: We know that M and N cannot both be on CY and CX respectively, or else the height from P to MN is longer than the height from Z to YX, which would make our supposition false, by Lemma 1. But I think the counterexample is easy to come up with Am I right? Dear pohoata, Could you post the solution involves an homothety and a translation? I think this problem is interesting and hard Thank you!

Ok, I will post it as soon as I will remember it (it has been a while).

Here is a different approach... Let $D,E,F$ be the midpoints of $BC,CA,AB$. Assume that $P,N$ lie on segments $AF,AE$, respectively. If $MH$ is the altitude from $M$ to $NP$ then $H$ lies on $NP$ and therefore $H$ lies inside triangle $AFE$, Therefore, if $MH \cap FE = H'$, then $MH > MH' $. Then the distance from $M$ to $NP$ is greater than the distance from $M$ to $FE$, which equals half of the distance from $A$ to $BC$ and we're done. Therefore we can assume $P \in AF, N \in EC, M \in BD$ (or $P \in BF, N \in EA, M \in DC$, but this is analogous). Fix triangle $ABC$, and consider the triangle right-angled or acute $MNP$ (satisfying the above condition) which minimizes $X$, and assume $X < x/2$. This is possible since we are taking $M,N,P$ in closed intervals. If, say, $M=D$, then notice the altitude from $P$ to $MN$ is greater than half of the altitude from $C$ to $AB$, so we're done. If, say, $M=B$, then the altitude from $N$ to $MP$ is greater than or equal to that of $E$ to $BA$ which is half of the altitude from $C$ to $AB$, so we're done. Otherwise, we can assume that $M,N,P$ are different from $A,B,C,D,E,F$. If we start to move $N$ towards $E$ (unless $ \angle N = 90$), then $X$ will increase, by definition, but the altitude from $P$ to $MN$ decreases and so does the altitude from $N$ to $MP$. Therefore, the altitude from $M$ to $PN$ must equal $X$ (otherwise, the resulting triangle $MN'P$ will have a smaller $X$, since the altitude from $M$ to $PN'$ increased but is still smaller than the original $X$). Similarly, the altitude from $N$ to $PM$ and the altitude from $P$ to $MN$ equal $X$. Therefore, $MNP$ is equilateral, or right-angled and isosceles. If it is right-angled and isosceles (with side $h$), say at $M$, notice that the distance from $P$ to $BC$ is at least half of that from $A$ to $BC$, and so it is $<h$. But it is also at least $PM = h$, contradiction! So proving it for $MNP$ equilateral is enough. Let $h$ be its altitude. Notice that, WLOG, $\angle A \ge 60$, and thus $A$ lies inside the reflection of the circumcircle of $MNP$ across $NP$. This implies that $A$ lies inside the circle centered at $M$ with radius $2h$ (since this circle engulfs the previous one). Thus, the altitude from $A$ to $BC$ is less than or equal to $AM$, which is less than or equal to $2h$. Thus, we're done.