Proving Altitudes of Triangle can never form a Triangle Prove that Altitudes of Triangle can never form a Triangle 
My try: we have altitudes proportional to reciprocals of sides of given triangle
Let $a,b,c$ are sides we have 
$$a+b \gt c $$
$$b+c \gt a$$ and $$c+a \gt b$$
Now if $\frac{1}{a}$, $\frac{1}{b}$ and $\frac{1}{c}$ are sides of another Triangle  we need to show that
$\frac{1}{a}+\frac{1}{b}$ Can never be Greater than $\frac{1}{c}$
Any hint?
 A: Let $a \geq b \geq c$.
Therefore, $\frac 1a \leq \frac 1b \leq \frac 1c$. 
The three form a triangle if and only if $\frac 1c < \frac 1b + \frac 1a$, which happens if and only if $c(a+b) > ab$. The altitudes are proportional to these quantities so they form a triangle if and only if line segments of these lengths do.
Which means that there may be triangles for which the altitudes do actually form a triangle , for example if $a=b=c$.
And there are examples where triangles are not formed e.g. if $c(a+b) \leq ab$. You can check that $3,3,1$ is a triangle, whose altitudes would not form a triangle since $1(3+3) = 6$ while $3 \times 3= 9$.
A: The altitudes of $\triangle ABC$ can form a triangle if and only if 

$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}>2\max\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)$$ 

where $a,b,c$ are the sides of $\triangle ABC.$
A: The altitudes can form a triangle for some but not all triangles $ABC$.
I. In isosceles triangle $ABC$, with altitudes $AD$, $BE$, $CF$, let $AC=2BC$.
Since$$\triangle ADC\sim\triangle BEC$$then$$\frac{AD}{BE}=\frac{AC}{BC}$$hence$$AD=2BE$$And since $BE=CF$, then $$AD=BE+CF$$and altitudes $AD$, $BE$, $CF$ cannot form a triangle when $\cos \angle ACB=\frac{DC}{AC}=.25$, that is for$$\angle ACB\ge\arccos .25=75.52^o$$But if $AC<2BC$, that is if $\angle ACB<75.52^O$, then since$$AD<2BE=BE+CF$$the three altitudes can form a triangle.
II. The altitudes of a right triangle can form a triangle only if the least angle is not much less than $30^o$.
In triangle $ABC$ with right angle at $C$, suppose $BC>AC$, and hence the least angle is at $B$. Points $E$, $D$ are coincident with $C$. Thus the altitudes can form a triangle only if$$AD+CF>BE$$
Constructing $FG\parallel CB$ and $\angle CBG=\angle FCB$, draw circle with radius $BG=CF$ making $BH=CF$. And let $AD$ be such that $AD=DH$. Then$$AD+CF=BE$$and the three altitudes cannot form a triangle. With $J$ the midpoint of $AB$, evidently $$AD<\frac12AB$$and hence$$\angle ABC<30^o$$
Then keeping $BD$ fixed it is clear that if we decrease $AD$, thereby also decreasing $CF$, then$$AD+CF<BE$$ and again the altitudes cannot form a triangle. But if instead we increase $AD$, hence also increasing $CF$, then$$AD+CF>BE$$and the altitudes can form a triangle. And they will continue to form a triangle until $AD$ has to $BE$ the ratio which $BE$ formerly had to $AD$, i.e. until the angles at $A$ and $B$ are the reverse of what they are in the figure. But if $AD$ is made greater still, the altitudes will no longer form a triangle.
Thus the altitudes of a right triangle can form a triangle if the complementary acute angles are between slightly greater than $60^o$, and slightly less than $30^o$.
It seems clear from just these two types of cases, that the altitudes of a triangle can form a triangle neither always nor never, but only under certain conditions. 
A: The condition for a triangle is
$c(a+b) > ab$.
For the 3-4-5 triangle,
this is
$5(3+4) > 3\cdot 4$
which is true.
So the altitudes do form a triangle
in this case.
Note that you can write
the condition as
$\begin{array}\\
0
&\gt ab-c(a+b)\\
&= ab-c(a+b)+c^2-c^2\\
&= (a-c)(b-c)-c^2\\
\text{or}\\
c^2
&\gt (a-c)(b-c)\\
&= (c-a)(c-b)\\
\end{array}
$
This is true if
$c$ is the largest side
of the triangle,
so it is always true.
Therefore,
the altitudes can
always form a triangle.
