inequality involving heights and bisectors Let $a,b,c,a \le b \le c$ be the sides of the triangle $ABC$, $l_a,l_b,l_c$ the lengths of its bisectors and $h_a,h_b,h_c$ the lengths of its heights. Prove that:
$$\frac {h_a+h_c} {h_b} \ge \frac {l_a+l_c} {l_b}$$
 A: Here's a proof.
The formulae for the lenghts of the angular bisectors and the heights are known. (see e.g. https://en.wikipedia.org/wiki/Triangle ). Let $T$ be the area of the triangle. We have
$$ l_a = \sqrt{bc (1 - \frac{a^2}{(b+c)^2})}$$
and
$$h_a = \frac{2 T}{a}$$
and cyclic shifts of those.
With these formulae, the required inequality gets
$$
\frac{1/a + 1/c}{1/b} \geq \frac{\sqrt{bc (1 - \frac{a^2}{(b+c)^2})}
+ \sqrt{ab (1 - \frac{c^2}{(a+b)^2})}}{\sqrt{ac (1 - \frac{b^2}{(a+c)^2})}}
$$
Multiplying  the RHS by 
$$1 = \frac{1/\sqrt{abc}}{1/\sqrt{abc}}
$$
gives 
$$
\frac{1/a + 1/c}{1/b} \geq \frac{\frac{1}{a}\sqrt{a (1 - \frac{a^2}{(b+c)^2})}
+ \frac{1}{c}\sqrt{c(1 - \frac{c^2}{(a+b)^2})}}{\frac{1}{b}\sqrt{b (1 - \frac{b^2}{(a+c)^2})}}
$$
Due to homogeneity, we can set $b=1$. This gives the condition $a\leq 1\leq c$ and the inequality gets
$$
\frac{1}{a} + \frac{1}{c} \geq \frac{1}{a}\sqrt{
a \frac{
1 - \frac{a^2}{(1+c)^2}
}{
1 - \frac{1}{(a+c)^2}
}
}
+ \frac{1}{c}\sqrt{
c\frac{
1 - \frac{c^2}{(a+1)^2}
}{
1 - \frac{1}{(a+c)^2}
}
}
$$
We now first prove that the first root is less or equal than 1, i.e. we need
$$
{a (1 - \frac{a^2}{(1+c)^2})} \leq 
1 - \frac{1}{(a+c)^2}
$$
which can be transformed into
$$
(a+c)^2 (1-a) \geq 1 - a^3 (\frac{a+c}{1+c})^2
$$
The last bracket  can be expanded and it suffices to prove 
$$
(a+c)^2 (1-a) \geq 1 - a^3 (1 - 2 \frac{1-a}{1+c})
$$
which shortens to
$$(1+c) a (1-a) \geq 2 a^3 (1-a)$$ or $$(1+c) \geq 2 a^2$$ which is certainly true under the given condition $a\leq 1\leq c$.
Likewise, we  prove that the second root is less or equal than 1, i.e. we need
$$
{c (1 - \frac{c^2}{(1+a)^2})} \leq 
1 - \frac{1}{(a+c)^2}
$$
which can be transformed into
$$
(a+c)^2 (1-c) \geq 1 - c^3 (\frac{a+c}{1+a})^2
$$
We can expand the last bracket and it suffices to prove 
$$
(a+c)^2 (1-c) \geq 1 - c^3  (1 + 2 \frac{c-1}{1+a})
$$
or in positive terms
$$
(a+c)^2 (c-1) \leq  c^3 -1 +   2 c^3 \frac{c-1}{1+a}$$
Since $a\leq1$, it suffices to prove
$$
(1+c)^2 (c-1)  - c^3 +1 \leq    2 c^3 \frac{c-1}{1+a}$$
or
$$
(1+a)(c^2 - c) \leq 2 c^2 (c^2-c)
$$
which shortens to
$$1+a \leq 2 c^2$$ which is certainly true under the given condition $a\leq 1\leq c$.
Hence both roots are less or equal to 1, which proves the inequality.
