Triangle inequality

Notice that $\frac { { h }_{ a } }{ a } \cdot \frac { { h }_{ b } }{ b } \cdot\frac { { h }_{ c } }{ c }=\sin { \alpha } \cdot \sin { \beta } \cdot \sin { \gamma }$ .

Since $\sin { \alpha } ,\sin { \beta } ,\sin { \gamma } >0$ we can use the AM-GM inequality to get that $\sin { \alpha } \cdot \sin { \beta } \cdot \sin { \gamma } \le { \left( \frac { \sin { \alpha } +\sin { \beta } +\sin { \gamma } }{ 3 } \right) }^{ 3 }$ .

Now since $f\left( x \right) =\sin { x }$ is concave on the interval $\left< 0,\pi \right>$ we can use Jensen's inequality to conclude that: ${ \left( \frac { \sin { \alpha } +\sin { \beta } +\sin { \gamma } }{ 3 } \right) }^{ 3 }\le { \left( \sin { \frac { \alpha +\beta +\gamma }{ 3 } } \right) }^{ 3 }={ \left( \sin { \frac { \pi }{ 3 } } \right) }^{ 3 }=\frac { 3\sqrt { 3 } }{ 8 } \approx 0.65$

Hence the answer is 0.65

try taking an equilateral triangle to solve.