It is enough to consider the inscribed triangle \(MNP\) for which its longest altitude is the minimal over all inscribed triangles. Then \(MNP\) is equilateral. Indeed, otherwise \(MNP\) has either one or two maximal altitudes. If there is one longest altitude and WLOG it comes out from \(M\) then we can twitch \(M\) so that this altitude is shortened while remaining the longest (if \(NP\parallel BC\) we can twitch one of \(N, P\)). Else, if there are two longest altitudes, WLOG coming out of \(M, N\) then we can again move \(M, N\) a bit towards \(C\) shortening these altitudes while keeping them equal in length. Thus the minimum of the longest altitude is achieved when all the altitudes are equal, i.e. \(MNP\) is regular. WLOG \(\angle A\ge 60^\circ\). Then if \(Q\) is the reflection of \(M\) in \(NP\) the point \(A\) lies inside the circle \(NPQ\) so \(\angle QAM\ge 90^\circ\) and \(h\le MA\le MQ=2H\) as needed.