Altitudes

True

Let the three altitudes be $a,b,c$ and their corresponding perpendicular sides be $x,y,z$ respectively. Then the area of triangle is given by: $\frac{1}{2}ax=\frac{1}{2}by=\frac{1}{2}cz$

Then $x=\dfrac{cz}{a},y=\dfrac{cz}{b}$

By triangle inequality, $\dfrac{cz}{a}+\dfrac{cz}{b} \gt z \implies c \gt \dfrac{ab}{a+b}$

Also, $\dfrac{cz}{a}+z \gt \dfrac{cz}{b} \implies c \lt \left|\dfrac{ab}{a-b}\right|$

So, $\boxed{\dfrac{ab}{a+b} \lt c \lt \left|\dfrac{ab}{a-b}\right|}$