Is it old result?

The condition of this problem is equivalent to the fact that $a$ , $b$ , $c$ can be the lengths of the sides of a triangle and $\frac 1 a$ , $\frac 1 b$ , $\frac 1 c$ can be the lengths of the sides of a triangle.

This a solution written in python 3.4:

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from random import random

def istriangle(a, b ,c ):
    return a + b > c and a + c > b and b + c > a

def solve(n):
    "Is it old result?"
    tot = 0
    for i in range(n):
        a, b, c =  random(),  random(), random()
        if min(a,b,c) > 0 and istriangle(a, b, c) and istriangle(1/a, 1/b, 1/c):
            tot += 1
    return tot/n

print(solve(10**7))

The angles of the initial triangle can be found using the law of cosines: $A = acos(\frac{a^2 - b^2 - c^2}{-2 \cdot b \cdot c})$ $B = acos(\frac{b^2 - a^2 - c^2}{-2 \cdot a \cdot c})$ $C = acos(\frac{c^2 - a^2 - b^2}{-2 \cdot a \cdot b})$

The altitudes are given (after simplification) by:

$h_a = c \cdot sin(B) = \frac{\sqrt{-(a - b - c) \cdot (a - b + c) \cdot (a + b - c) \cdot (a + b + c)}}{2 \cdot a}$ $h_b = a \cdot sin(C) = \frac{\sqrt{-(a - b - c) \cdot (a - b + c) \cdot (a + b - c) \cdot (a + b + c)}}{2 \cdot b}$ $h_c = b \cdot sin(A) = \frac{\sqrt{-(a - b - c) \cdot (a - b + c) \cdot (a + b - c) \cdot (a + b + c)}}{2 \cdot c}$

Now in order to satisfy the triangle inequality and the given constraints the following must hold:

$0 < a < 1; 0 < b < 1; 0 < c < 1$ $a + b > c; a + c > b; b + c > a$ $h_a + h_b > h_c; h_a + h_c > h_b; h_b + h_c > h_a$

The last set of inequalities can be written as

$c \cdot (a + b) > a \cdot b$ $a \cdot (b + c) > b \cdot c$ $b \cdot (a + c) > a \cdot c$

I then used Mathematica to solve these inequalities to get 11 different regions which could be integrated over to get a final answer of

$\frac{1}{6} \cdot (6-3 \cdot \sqrt{5}+log(4096)+log(161-72 \cdot \sqrt{5})) \approx 0.30583672$