Triangle inequalities help

I got it! The answer is $({4\over5},{4\over3})$.

Here is how I got it:

Let S be the twice the area of the triangle. Then the side lengths are S, S/2, and S/h. Since the triangle is acute, these side lengths must satisfy these inequalities:

$S^2 + ({S\over2})^2>({S\over h})^2$

$({S\over2})^2+({S\over h})^2 >S^2$

$S^2 + ({S\over h})^2>({S\over2})^2$

We can divide each inequality by $S^2$ to simplify, and the third inequality will automatically be satisfied, so we are left with:

$1 + {1\over4} > {1\over h^2}$

${1\over4} + {1\over h^2} > 1$

Simplifying, we get h^2>4/5, and h^2<4/3, so the interval for h^2 is the (4/5, 4/3).