Heights and sides form two triangles 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?
 A: 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
