Given an acute triangle, consider the three sides and the three heights. Show that using these six segments, it is possible to form two triangles.
For three segments with length $a,b,c$ to form a triangle, we need $a+b>c,b+c>a,c+a>b$. The heights of the triagle are $2A/a,2A/b,2A/c$, where $A=\sqrt{s(s-a)(s-b)(s-c)}$ is the area of the triangle and $s=(a+b+c)/2$.
If $a\geq b\geq c$, then $2A/a\leq 2A/b\leq 2A/c$. To check the triangle condition is equivalent to checking $2A/a+2A/b>c$ for instance, but this looks rather involved. Is there a way to get around it?
This is a "simple in principle" but perhaps "long-winded in practice" way to show this.
Assume an acute triangle with sides of length $a$, $b$ and $c$, where $a \ge b \ge c$, and with angles $A$, $B$ and $C$ (see figure below). Denote the height from the vertex with angle $A$ as $h_A$, the height from the vertex with angle $B$ as $h_B$, etc.
Contention: the segments of length $a$, $b$ and $h_C$ can always make a triangle and the segments of length $c$, $h_A$ and $h_B$ can always make a triangle.
Proof
Some initial deductions will be done and then, using the results of these deductions and the triangle inequality, the final proof will de done in section $4$ below.
$1$. Formulas for the heights
The area of the triangle can be found as
$Area = \frac{1}{2}ab\,sin(C) =\frac{1}{2}ac\,sin(B) =\frac{1}{2}bc\,sin(A)$
or as
$Area = \frac{1}{2}a\,h_A =\frac{1}{2}b\,h_B = \frac{1}{2}c\,h_C$
giving
$h_A = b\,sin(C) = c\,sin(B)$
$h_B = a\,sin(C) = c\,sin(A)$
$h_C = a\,sin(B) = b\,sin(A)$
$2$. Limits of the angles
For any triangle where the side lengths are ordered as $a \ge b \ge c$, the associated angle opposite each side will likewise be ordered as $A \ge B \ge C$ (can be shown using Law of Sines).
For an acute triangle where all angles are less than $90^{\circ}$, the following limits to the angles can be deduced:
max($A$) $\lt 90^{\circ}$ and min($A$) $= 60^{\circ}$
max($B$) $\lt 90^{\circ}$ and min($B$) $\gt 45^{\circ}$
max($C$) $= 60^{\circ}$ and min($C$) $\gt 0^{\circ}$
As an example, let's look at angle $B$. For the maximum, we see that angle $B$ can at most be equal to angle $A$ and angle $A$ can at most be just under $90^{\circ}$. As the sum of the $3$ angles must be $= 180^{\circ}$, having $2$ angles just under $90^{\circ}$ let's angle $C$ be just over $0^{\circ}$, which is fine. For the minimum, angles $A$ and $C$ must be as large as possible. Angle $A$ can at most be just under $90^{\circ}$, leaving angles $B$ and $C$ to share just over $90^{\circ}$. Dividing equally, angle $B$ must be at least $45^{\circ}$.
$3$. Limits for the heights
Using the formulas for the heights found in section $1$ and the maximum and minimum values of the angles found in section $2$ and the fact that
Sin$(0^{\circ}) = 0$
Sin$(45^{\circ}) = \frac{1}{\sqrt 2}$
Sin$(60^{\circ}) = \frac{\sqrt 3}{2}$
Sin$(90^{\circ}) = 1$
we can deduce the following limits to the heights: $$h_A:\quad c \gt h_A \gt \frac{c}{\sqrt 2} \quad , \quad \frac{\sqrt 3}{2}b \gt h_A \gt 0$$ $$h_B:\quad c \gt h_B \gt \frac{\sqrt 3}{2}c \quad , \quad \frac{\sqrt 3}{2}a \gt h_B \gt 0$$ $$h_C:\quad a \gt h_C \gt \frac{a}{\sqrt 2} \quad , \quad b \gt h_C \gt \frac{\sqrt 3}{2}b$$
As an example, let's look at $h_C$. We know that $$h_C = a\,sin(B)$$
Plugging in the maximum value of $B$ (max($B$) $\lt 90^{\circ}$) we find that $$h_C \lt a$$
Plugging in the minimum value of $B$ (min($B$) $\gt 45^{\circ}$) we find that $$h_C \gt \frac{a}{\sqrt 2} $$ These two inequalities combined gives the first inequality for $h_C$ given above. The second inequality for $h_C$ is found by using the formula $h_C = b\,sin(A)$ and plugging in the min and max values for $A$. Likewise for the rest.
$4$. Using the triangle inequality to prove the contention
The contention was "the segments of length $a$, $b$ and $h_C$ can always make a triangle and the segments of length $c$, $h_A$ and $h_B$ can always make a triangle". Both parts of this contention will be proven using the triangle inequality.
First part
The segments of length $a$, $b$ and $h_C$ can always make a triangle iff the following is true $$a+b \gt h_C \quad \land \quad a+h_C \gt b \quad \land \quad b+h_C \gt a$$
$4a. \;$Proving that $\; a+b \gt h_C$
We know from section $3$ of the proof that $a \gt h_C$ and $b \gt h_C$ and therefore $$a+b \gt h_C$$
$4b. \;$Proving that $\; a+h_C \gt b$
We know from the initial assumptions that $a \ge b$ and we know that $h_C \gt 0$ and therefore $$a+h_C \gt b$$
$4c. \;$Proving that $\; b+h_C \gt a$
The triangle inequality must hold for our given triangle and it must therefore be true that $b+c \gt a$. From the initial assumptions we know that $b \ge c$. Therefore $$b+c \gt a$$ $$=> \; b+b \gt a$$ $$=> \; b \gt \frac{a}{2}$$
From section $3$ we also know that $$h_C \gt \frac{a}{\sqrt 2}$$ $$=> \; b+h_C \gt \frac{a}{2}+\frac{a}{\sqrt 2}$$ $$=> \; b+h_C \gt a$$
The first part of the contention has thus been demonstrated.
Second part
The segments of length $c$, $h_A$ and $h_B$ can always make a triangle iff the following is true $$c+h_A \gt h_B \quad \land \quad c+h_B \gt h_A \quad \land \quad h_A+h_B \gt c$$
$4d. \;$Proving that $\; c+h_A \gt h_B$
We know from section $3$ of the proof that $c \gt h_B$ and we know that $h_A \gt 0$ and therefore $$c+h_A \gt h_B$$
$4e. \;$Proving that $\; c+h_B \gt h_A$
We know from section $3$ of the proof that $c \gt h_A$ and therefore $$c+h_B \gt h_A$$
$4f. \;$Proving that $\; h_A+h_B \gt c$
From section $3$ of the proof we know that $h_A \gt \frac{c}{\sqrt 2}$. We also know that $h_B \gt \frac{\sqrt 3}{2}c$. This means that $$h_A + h_B \gt \frac{c}{\sqrt 2}+ \frac{\sqrt 3}{2}c$$ $$=> h_A + h_B \gt c$$
The second part of the contention has thus been demonstrated. As both parts of the contention have now been proven:
QED