Does there exist a triangle with altitudes $h_a,h_b,$ and $h_c$ for all $h_a,h_b,h_c\in\Bbb{R}^{+}$? Here's my attempt.
If we have a triangle with side lengths $a,b,$ and $c$ and altitudes with length $h_a,h_b$ and $h_c$, then these must satisfy the following, where $s=\frac{1}{2}(a+b+c)$:

*

*$h_a=\frac{2\sqrt{s(s-a)(s-b)(s-c)}}{a}$

*$h_b=\frac{2\sqrt{s(s-a)(s-b)(s-c)}}{b}$

*$h_c=\frac{2\sqrt{s(s-a)(s-b)(s-c)}}{c}$
With a bit of reworking, we can say

*

*$(ah_a)^2=4s(s-a)(s-b)(s-c)$

*$(bh_b)^2=4s(s-a)(s-b)(s-c)$

*$(ch_c)^2=4s(s-a)(s-b)(s-c)$
But honestly, I think this might be even more of a headache to work with in general.
I will say that the motivating factor here was whether or not you can have a triangle with altitudes of 2, 3, and 6 (which would make the incircle's area a nice, clean $\pi$ units squared)m and for that case, we have

*

*$a^2=s(s-a)(s-b)(s-c)$

*$9b^2=4s(s-a)(s-b)(s-c)$

*$9c^2=s(s-a)(s-b)(s-c)$
which might be a bit easier than the general form, but it still seems intimidating to me. As a system of three equations with three unknowns, I feel like you should be able to solve for it, but I also know that we're dealing with some rather messy quartic equations well, so that could throw a wrench in things. As always, help is appreciated. Thank you!
 A: Use the fact that each side is twice the area of the triangle $K$ divided by the corresponding altitude. Thus from the Triangle Inequality:
$(2K/h_a)\le(2K/h_b)+(2K/h_c)$
$\color{blue}{(1/h_a)\le(1/h_b)+(1/h_c)}$
Thus the minimum altitude must satisfy this inequality to have a real figure, and if the inequality is saturated the triangle is degenerated.
In your case $(1/2)=(1/3)+(1/6)$, so your triangle is degenerated.
To get an incitcle of unit radius try a $3-4-5$ right triangle whose respective altitudes are $4,3,2.4$.
Upside down
Suppose you were to take the familiar formula for the area of a triangle in termsbof its sides
$K=\sqrt{s(s-a)(s-b)(s-c)}$
with the semiperimeter $s=(a+b+c)/2$ and render each side as twice the area divided by the altitude ($a=2K/h_a$, etc). Re-solving for the area gives a reciprocal formula in terms of the altitudes of the triangle:
$\frac{1}{K}=\sqrt{\left(\frac{1}{r}\right)\left(\frac{1}{r}-\frac{2}{h_a}\right)\left(\frac{1}{r}-\frac{2}{h_b}\right)\left(\frac{1}{r}-\frac{2}{h_c}\right)}$
where $r$ is the radius of the inscribed circle given by
$1/r=(1/h_a)+(1/h_b)+(1/h_c).$
This says the triangle has a real, finite area if the diameter of the inscribed circle is less than each of the altitudes. Given the radius determination above, this condition for a finite sized, nondegenerate triangle is equivalent to the reciprocal of each altitude being less than the sum of the other two reciprocal altitudes.
Thus an incircle diameter of $2$ coupled with altitudes of $2.4,3,4$ will give a triangle (the $3-4-5$ right triangle cited above), but the same diameter coupled with altitudes of $2,3,6$ will fail to give a nondegenerate triangle.
A: Consider a triangle with side lengths $a, b, c$ and altitudes $h_a, h_b, h_c$.
We have
$$\text{Area} = \frac{1}{2} \times a \times h_a = \frac{1}{2} \times b \times h_b = \frac{1}{2} \times c \times h_c$$
$$\implies ah_a = bh_b = ch_c$$
$$\therefore a = \frac{1}{h_a}k, b = \frac{1}{h_b}k, c = \frac{1}{h_c}k$$
Without loss of generality, we can assume that $0 < a \leq b \leq c$.
It follows from the triangle inequality that the triangle exists iff
$$\frac{1}{h_a}k + \frac{1}{h_b}k \geq \frac{1}{h_c}k \iff \frac{1}{h_a} + \frac{1}{h_b} \geq \frac{1}{h_c}$$
